Class 10 Exercise 1.3 Solutions — Complex Numbers

Concepts

Definition

The complex plane

A complex number $z=x+iy$ is fixed by the ordered pair of real numbers $(x,y)$ , so it can be drawn as the point $(x,y)$ in a plane. The horizontal axis carries the real part and is called the real axis. The vertical axis carries the imaginary part and is called the imaginary axis. This plane is called the complex plane, or the Argand plane.

A complex number can also be drawn as an arrow from the origin to that point. The number, the point and the arrow are all written with the same letter $z$ .

The plane is named after Jean Robert Argand $(1768$ to $1822)$ , a Swiss mathematician.

Examples

$2-5i$ is the point $(2,-5)$ .
$8$ is the point $(8,0)$ , which lies on the real axis, because its imaginary part is $0$ .
$-7i$ is the point $(0,-7)$ , which lies on the imaginary axis, because its real part is $0$ .

Definition

Conjugate of a complex number

The conjugate of $z=x+iy$ is written $\bar{z}$ and is defined by $\bar{z}=x-iy$ In words: keep the real part exactly as it is, and change the sign of the imaginary part.

In rectangular form with real coefficients, replace $x+iy$ by $x-iy$ . The real part stays unchanged.

Conjugation reflects a point across the real axis. A point above the axis moves the same distance below it, and conversely. A point on the real axis stays where it is.

The product of a complex number with its conjugate is always a real number. This is what makes the conjugate useful for division.

Examples

The conjugate of $3+8i$ is $3-8i$ .
The conjugate of $-5-4i$ is $-5+4i$ .
The conjugate of $6$ is $6$ , because its imaginary part is already $0$ .
$(1+2i)(1-2i)=1^{2}-(2i)^{2}=1+4=5$ , a real number.

Rule

Properties of the conjugate

For complex numbers $z$ , $z_{1}$ , and $z_{2}$ , $\begin{aligned}\overline{z_{1}+z_{2}}&=\bar z_{1}+\bar z_{2}, \\ \overline{z_{1}z_{2}}&=\bar z_{1}\bar z_{2}, \\ \overline{\left(\frac{z_{1}}{z_{2}}\right)}&=\frac{\bar z_{1}}{\bar z_{2}}\quad(z_{2}\neq0), \\ \overline{\bar z}&=z\end{aligned}$

In words: taking the conjugate can be done before or after adding, multiplying or dividing, and the answer is the same either way.

The last one says that taking the conjugate twice brings you back to where you started. Reflecting in the real axis twice returns the point to its original place.

Examples

For $z_{1}=4+3i$ and $z_{2}=5+2i$ : $\ \overline{z_{1}+z_{2}}=\overline{9+5i}=9-5i$ , and $\bar{z_{1}}+\bar{z_{2}}=(4-3i)+(5-2i)=9-5i$ .
For $z=7+3i$ : $\ \bar{z}=7-3i$ , so $\bar{\bar{z}}=7+3i=z$ .

Definition

Modulus of a complex number

The modulus of $z=x+iy$ is written $|z|$ and is defined by $|z|=\sqrt{x^{2}+y^{2}}$

The modulus is the distance from the origin to the point $(x,y)$ . The horizontal and vertical distances are $|x|$ and $|y|$ , not possibly negative coordinates. Pythagoras gives $|z|^{2}=x^{2}+y^{2}$ ; the formula also holds on either axis and at the origin.

A modulus is a nonnegative real number, and $|z|=0$ if and only if $z=0$ .

Square each part before adding. Note that $|3-4i|$ is $\sqrt{9+16}=5$ , not $3-4$ .

For a real number this is just the size without the sign, which is why the same two bars are used.

Examples

$|i|=|0+1i|=\sqrt{0^{2}+1^{2}}=1$
$|3-5i|=\sqrt{3^{2}+(-5)^{2}}=\sqrt{9+25}=\sqrt{34}$
$|{-6}|=\sqrt{(-6)^{2}+0^{2}}=6$

Rule

Properties of the modulus

For any complex number $z=x+iy$ , $|z|=|\bar{z}|=|{-z}|=|{-\bar{z}}|=\left|\bar{\bar{z}}\right|, \qquad z\bar{z}=|z|^{2}$

The first chain says that changing signs does not change distance. Every one of $z$ , $\bar{z}$ , $-z$ and $-\bar{z}$ has the same two numbers being squared, so the sum $x^{2}+y^{2}$ is the same for all of them.

The identity $z\bar z=|z|^{2}$ explains why multiplication by the conjugate produces a real denominator. This denominator is strictly positive when $z\neq0$ .

Note carefully that $z\bar{z}$ gives $|z|^{2}$ , not $|z|$ . There is no square root at the end.

Examples

For $z=5+4i$ : $\ |z|=\sqrt{41}$ and $|\bar{z}|=|5-4i|=\sqrt{25+16}=\sqrt{41}$ .
For $z=5+3i$ : $\ z\bar{z}=(5+3i)(5-3i)=25+9=34$ , and $|z|^{2}=\left(\sqrt{34}\right)^{2}=34$ .

Deep understanding

Why conjugation preserves sums

Let $z_{1}=a+bi$ and $z_{2}=c+di$ , where $a,b,c,d$ are real. Add first and then conjugate:

$\begin{aligned}\overline{z_{1}+z_{2}}&=\overline{(a+c)+(b+d)i} \\ &=(a+c)-(b+d)i\end{aligned}$

Conjugating first and then adding gives the same expression:

$\begin{aligned}\bar z_{1}+\bar z_{2}&=(a-bi)+(c-di) \\ &=(a+c)-(b+d)i\end{aligned}$

Thus $\overline{z_{1}+z_{2}}=\bar z_{1}+\bar z_{2}$ for every pair of complex numbers, not just for one numerical example.

Why conjugation preserves products

For the same real coordinates, multiplication gives $z_{1}z_{2}=(ac-bd)+(ad+bc)i$ . Consequently,

$\begin{aligned}\overline{z_{1}z_{2}}&=(ac-bd)-(ad+bc)i\end{aligned}$

Now multiply the conjugates directly:

$\begin{aligned}\bar z_{1}\bar z_{2}&=(a-bi)(c-di) \\ &=ac-adi-bci+bd\,i^{2} \\ &=(ac-bd)-(ad+bc)i\end{aligned}$

The two results agree, so $\overline{z_{1}z_{2}}=\bar z_{1}\bar z_{2}$ .

Why conjugation preserves quotients

Assume $z_{2}\neq0$ and let $w=z_{1}/z_{2}$ . Then $wz_{2}=z_{1}$ . Conjugate both sides and use the product property just proved:

$\begin{aligned}\overline{wz_{2}}&=\bar z_{1}, \\ \bar w\,\bar z_{2}&=\bar z_{1}\end{aligned}$

The conjugate $\bar z_{2}$ is nonzero: if it were $0$ , both real coordinates of $z_{2}$ would be $0$ . Divide by $\bar z_{2}$ to obtain

$\begin{aligned}\overline{\left(\frac{z_{1}}{z_{2}}\right)}&=\frac{\bar z_{1}}{\bar z_{2}}\end{aligned}$

Why conjugating twice returns the number

If $z=x+iy$ with $x,y$ real, its conjugate is $x-iy=x+i(-y)$ . A second conjugation changes $-y$ back to $y$ :

$\begin{aligned}\overline{\bar z}&=x-i(-y) \\ &=x+iy \\ &=z\end{aligned}$

Why the modulus is a distance

Draw $z=x+iy$ as the point $P(x,y)$ . Draw a perpendicular from $P$ to the real axis, meeting it at $M=(x,0)$ . Depending on the sign of $y$ , this segment goes upward or downward.

When $x$ and $y$ are both nonzero, $OMP$ is a right triangle. Its horizontal and vertical side lengths are $|x|$ and $|y|$ , not possibly negative coordinates. Since $|x|^{2}=x^{2}$ and $|y|^{2}=y^{2}$ , Pythagoras gives the formula below. If either coordinate is $0$ , the same formula gives the distance along an axis.

For the nondegenerate triangle, Pythagoras gives $\begin{aligned}OP^{2}&=OM^{2}+MP^{2} \\ &=|x|^{2}+|y|^{2} \\ &=x^{2}+y^{2}\end{aligned}$

Take the square root of both sides, and take the nonnegative root, because a distance is never negative. $\begin{aligned}OP &= \sqrt{x^{2}+y^{2}} = |z|\end{aligned}$

At the origin, both coordinates and the distance are $0$ . Thus the same formula covers every point in the complex plane.

Why $|z|$ , $|\bar{z}|$ , $|{-z}|$ and $|{-\bar{z}}|$ are all equal

Write the four numbers out side by side and see what changes. $\begin{aligned}z &= x+iy \\ \bar{z} &= x-iy \\ -z &= -x-iy \\ -\bar{z} &= -x+iy\end{aligned}$

Only the signs change. The real part is $x$ or $-x$ , and the imaginary part is $y$ or $-y$ .

Now put each one into the modulus formula. Squaring destroys a minus sign, because minus times minus gives plus. $\begin{aligned}(-x)^{2} &= x^{2} \\ (-y)^{2} &= y^{2}\end{aligned}$

So every one of the four gives the same sum inside the root, namely $x^{2}+y^{2}$ , and therefore the same modulus.

The complex plane gives a geometric explanation. Reflecting a point in an axis, or turning it through $180^\circ$ about the origin, never changes how far it is from the origin.

Why $z\bar{z}=|z|^{2}$

Take $z=x+iy$ , so that $\bar{z}=x-iy$ , and multiply them. The two brackets are of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ , with $a=x$ and $b=iy$ . $\begin{aligned}z\bar{z} &= (x+iy)(x-iy) \\ &= x^{2}-(iy)^{2}\end{aligned}$

Now open the second square. Square both the $i$ and the $y$ . $\begin{aligned}(iy)^{2} &= i^{2}y^{2} \\ &= (-1)y^{2} = -y^{2}\end{aligned}$

Subtract it. Taking away $-y^{2}$ is the same as adding $y^{2}$ . $\begin{aligned}z\bar{z} &= x^{2}-(-y^{2}) \\ &= x^{2}+y^{2}\end{aligned}$

But $x^{2}+y^{2}$ is exactly what sits inside the square root in the modulus, so it is $|z|^{2}$ . $\begin{aligned}|z|^{2} &= \left(\sqrt{x^{2}+y^{2}}\right)^{2} = x^{2}+y^{2}\end{aligned}$

So $z\bar{z}=|z|^{2}$ . Two things follow. The product is always real, which is why the conjugate clears a denominator. And it is never negative, which fits a squared distance.

The Argand diagram

Caspar Wessel published a geometric interpretation of complex numbers in 1799. Jean-Robert Argand developed an independent account in 1806; the name Argand diagram is commonly used for the complex plane.

The same diagram connects the algebraic ideas in this unit: conjugation is reflection across the real axis, negation is a half-turn about the origin, and modulus is distance from the origin.

Examples

Example 11

If $z_{1}=4+3i$ and $z_{2}=5+2i$ , prove that $\overline{z_{1}+z_{2}}=\bar{z_{1}}+\bar{z_{2}}$ .

Solution

Compact solution

To prove an equality like this, work out each side on its own and show they agree. Start with the left side, which says add first, then take the conjugate. So add the two numbers.
$z_{1}+z_{2} = (4+3i)+(5+2i) = (4+5)+(3+2)i = 9+5i$
Now take the conjugate of that sum. Keep the real part and change the sign of the imaginary part.
$\overline{z_{1}+z_{2}} = \overline{9+5i} = 9-5i \qquad \cdots(1)$
Now the right side, which says take the conjugates first, then add. Take each conjugate on its own.
$\bar{z_{1}} = 4-3i, \qquad \bar{z_{2}} = 5-2i$
Add them, real parts together and imaginary parts together.
$\bar{z_{1}}+\bar{z_{2}} = (4+5)+(-3-2)i = 9-5i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides give the same number, which is what was to be proved.
$\overline{z_{1}+z_{2}} = \bar{z_{1}}+\bar{z_{2}} = 9-5i$

AnswerBoth sides give

9-5i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}z_1+z_2 &= (4+3i)+(5+2i) \\ &= 4+3i+5+2i \\ &= (4+5)+(3+2)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 9+5i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{z_1+z_2} &= \overline{9+5i} \\ &= 9-5i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=4-3i \\ \bar z_2=5-2i\end{gathered}

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}\bar z_1+\bar z_2 &= (4-3i)+(5-2i) \\ &= 4-3i+5-2i \\ &= (4+5)+(-3-2)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 9-5i\end{aligned}

Both sides give $9-5i$ , so the required equality holds for these values.

\overline{z_1+z_2}=\bar z_1+\bar z_2=9-5i

Example 12

If $z_{1}=3+4i$ and $z_{2}=2+3i$ , prove that $\overline{z_{1}z_{2}}=\bar{z_{1}}\,\bar{z_{2}}$ .

Solution

Compact solution

Work out each side on its own and show they agree. The left side says multiply first, then take the conjugate. So multiply the brackets out term by term.
$z_{1}z_{2} = (3+4i)(2+3i) = 6+9i+8i+12i^{2}$
Replace $i^{2}$ by $-1$ , so the last term becomes $-12$ .
$12i^{2} = 12(-1) = -12$
Collect. The real numbers give $6-12$ , and the $i$ terms give $9+8$ .
$z_{1}z_{2} = (6-12)+(9+8)i = -6+17i$
Now take the conjugate of that product, changing the sign of the imaginary part.
$\overline{z_{1}z_{2}} = -6-17i \qquad\cdots(1)$
Turn to the right side, which says take the conjugates first. Write each one, keeping the real part and changing the sign of the imaginary part.
$\bar{z_{1}} = 3-4i, \qquad \bar{z_{2}} = 2-3i$
Multiply them out term by term.
$\bar{z_{1}}\,\bar{z_{2}} = (3-4i)(2-3i) = 6-9i-8i+12i^{2}$
Replace $i^{2}$ by $-1$ again.
$12i^{2} = -12$
Collect. Note that $-9-8=-17$ , because going down $8$ more from $-9$ reaches $-17$ .
$\bar{z_{1}}\,\bar{z_{2}} = (6-12)+(-9-8)i = -6-17i \qquad\cdots(2)$
From $(1)$ and $(2)$ both sides give the same number, which is what was to be proved.
$\overline{z_{1}z_{2}} = \bar{z_{1}}\,\bar{z_{2}} = -6-17i$

AnswerBoth sides give

-6-17i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Multiply each term in the first bracket by each term in the second.

\begin{aligned}z_1z_2 &= (3+4i)(2+3i) \\ &= 3(2)+3(3i)+(4i)(2)+(4i)(3i) \\ &= 6+9i+8i+12i^2 \\ &= 6+9i+8i+12(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 6+9i+8i-12 \\ &= (6-12)+(9+8)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -6+17i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{z_1z_2} &= \overline{-6+17i} \\ &= -6-17i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=3-4i \\ \bar z_2=2-3i\end{gathered}

Multiply each term in the first bracket by each term in the second.

\begin{aligned}\bar z_1\bar z_2 &= (3-4i)(2-3i) \\ &= 3(2)+3(-3i)+(-4i)(2)+(-4i)(-3i) \\ &= 6-9i-8i+12i^2 \\ &= 6-9i-8i+12(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 6-9i-8i-12 \\ &= (6-12)+(-9-8)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -6-17i\end{aligned}

Both sides give $-6-17i$ , so the required equality holds for these values.

\overline{z_1z_2}=\bar z_1\bar z_2=-6-17i

Example 13

If $z_{1}=5+4i$ and $z_{2}=3+2i$ , prove that $\overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\bar{z_{1}}}{\bar{z_{2}}}$ .

Solution

Compact solution

Take the left side first. It says divide first, then take the conjugate. So begin with the division.
$\frac{z_{1}}{z_{2}} = \frac{5+4i}{3+2i}$
There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate of the bottom, which is $3-2i$ .
$\frac{5+4i}{3+2i} = \frac{5+4i}{3+2i}\times\frac{3-2i}{3-2i}$
Work out the numerator, multiplying term by term.
$(5+4i)(3-2i) = 15-10i+12i-8i^{2}$
Replace $i^{2}$ by $-1$ . Note that $-8(-1)=+8$ , because minus times minus gives plus.
$-8i^{2} = 8$
Collect the numerator.
$(5+4i)(3-2i) = (15+8)+(-10+12)i = 23+2i$
Now the denominator, using $(a+b)(a-b)=a^{2}-b^{2}$ .
$(3+2i)(3-2i) = 3^{2}-(2i)^{2} = 9-4i^{2} = 9+4 = 13$
Put the numerator over the denominator and split into two parts.
$\frac{z_{1}}{z_{2}} = \frac{23+2i}{13} = \frac{23}{13}+\frac{2}{13}i$
Now take the conjugate of that, changing the sign of the imaginary part.
$\overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{23}{13}-\frac{2}{13}i \qquad \cdots(1)$
Turn to the right side, which says take the conjugates first, then divide.
$\bar{z_{1}} = 5-4i, \qquad \bar{z_{2}} = 3-2i$
Divide them, again multiplying top and bottom by the conjugate of the new bottom, which is $3+2i$ .
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{5-4i}{3-2i}\times\frac{3+2i}{3+2i}$
Work out the numerator.
$(5-4i)(3+2i) = 15+10i-12i-8i^{2}$
Replace $i^{2}$ by $-1$ and collect.
$(5-4i)(3+2i) = (15+8)+(10-12)i = 23-2i$
The denominator is the same as before.
$(3-2i)(3+2i) = 9+4 = 13$
So the right side comes to this.
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{23-2i}{13} = \frac{23}{13}-\frac{2}{13}i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides agree, and this is the value they share.
$\overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{23}{13}-\frac{2}{13}i$

AnswerBoth sides give

\dfrac{23}{13}-\dfrac{2}{13}i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Multiply numerator and denominator by the conjugate $3-2i$ .

\begin{aligned}\frac{z_1}{z_2} &= \frac{5+4i}{3+2i} \\ &= \frac{5+4i}{3+2i}\cdot\frac{3-2i}{3-2i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{3-2i}{3-2i}=1$}} \\ &= \frac{(5+4i)(3-2i)}{(3+2i)(3-2i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(5+4i)(3-2i) &= 5(3)+5(-2i)+(4i)(3)+(4i)(-2i) \\ &= 15-10i+12i-8i^2 \\ &= 15-10i+12i-8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15-10i+12i+8 \\ &= (15+8)+(-10+12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 23+2i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(3+2i)(3-2i) &= 3^2-(2i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 9-4i^2 \\ &= 9-4(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 9+4 \\ &= 13\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{z_1}{z_2} &= \frac{23+2i}{13} \\ &= \frac{23}{13}+\frac{2}{13}i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{\frac{z_1}{z_2}} &= \overline{\dfrac{23}{13}+\dfrac{2}{13}i} \\ &= \dfrac{23}{13}-\dfrac{2}{13}i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=5-4i \\ \bar z_2=3-2i\end{gathered}

Multiply numerator and denominator by the conjugate $3+2i$ .

\begin{aligned}\frac{\bar z_1}{\bar z_2} &= \frac{5-4i}{3-2i} \\ &= \frac{5-4i}{3-2i}\cdot\frac{3+2i}{3+2i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{3+2i}{3+2i}=1$}} \\ &= \frac{(5-4i)(3+2i)}{(3-2i)(3+2i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(5-4i)(3+2i) &= 5(3)+5(2i)+(-4i)(3)+(-4i)(2i) \\ &= 15+10i-12i-8i^2 \\ &= 15+10i-12i-8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15+10i-12i+8 \\ &= (15+8)+(10-12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 23-2i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(3-2i)(3+2i) &= 3^2-(2i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 9-4i^2 \\ &= 9-4(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 9+4 \\ &= 13\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{\bar z_1}{\bar z_2} &= \frac{23-2i}{13} \\ &= \frac{23}{13}-\frac{2}{13}i\end{aligned}

Both sides give $\dfrac{23}{13}-\dfrac{2}{13}i$ , so the required equality holds for these values.

\overline{\frac{z_1}{z_2}}=\frac{\bar z_1}{\bar z_2}=\dfrac{23}{13}-\dfrac{2}{13}i

Example 14

If $z=7+3i$ , prove that $\bar{\bar{z}}=z$ .

Solution

Compact solution

Take the conjugate once. Keep the real part and change the sign of the imaginary part.
$\bar{z} = 7-3i$
Now take the conjugate again, of the number you have just found. The real part $7$ stays, and the imaginary part $-3$ changes sign to $+3$ .
$\bar{\bar{z}} = \overline{7-3i} = 7+3i$
That is the number we started with, which is what was to be proved.
$\bar{\bar{z}} = 7+3i = z$

Answer

\bar{\bar{z}}=7+3i=z

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{7+3i} \\ &= 7-3i\end{aligned}

Conjugate once more; the imaginary sign changes back.

\begin{aligned}\overline{\bar{z}} &= \overline{7-3i} \\ &= 7+3i \\ &= z\end{aligned}

Common mistake

Watch outOn the complex plane this says that reflecting a point in the real axis twice brings it back to where it started.

Example 15

If $z=5+4i$ , show that $|z|=|\bar{z}|$ .

Solution

Compact solution

The modulus of $z=x+iy$ is $\sqrt{x^{2}+y^{2}}$ . Here $x=5$ and $y=4$ .
$|z| = \sqrt{5^{2}+4^{2}}$
Square each part.
$5^{2} = 25, \qquad 4^{2} = 16$
Add them, then take the square root. Since $41$ is not a perfect square, the root stays as it is.
$|z| = \sqrt{25+16} = \sqrt{41} \qquad \cdots(1)$
Now write the conjugate, changing the sign of the imaginary part.
$\bar{z} = 5-4i$
Take its modulus. Here $x=5$ and $y=-4$ .
$|\bar{z}| = \sqrt{5^{2}+(-4)^{2}}$
Square each part. Note that $(-4)^{2}=16$ , because minus times minus gives plus, so the minus sign disappears.
$(-4)^{2} = 16$
Add and take the root.
$|\bar{z}| = \sqrt{25+16} = \sqrt{41} \qquad \cdots(2)$
From $(1)$ and $(2)$ both are the same, which is what was to be shown.
$|z| = |\bar{z}| = \sqrt{41}$

AnswerBoth sides give

\sqrt{41}

, so the property holds.

Compact solution

The real and imaginary coefficients are $5$ and $4$ . Use the nonnegative square root.

\begin{aligned}|z| &= \sqrt{5^2+4^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+16} \\ &= \sqrt{41}\end{aligned}

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{5+4i} \\ &= 5-4i\end{aligned}

The real and imaginary coefficients are $5$ and $-4$ . Use the nonnegative square root.

\begin{aligned}\left|\bar{z}\right| &= \sqrt{5^2+(-4)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+16} \\ &= \sqrt{41}\end{aligned}

The two distances are equal.

|z|=\left|\bar{z}\right|=\sqrt{41}

Example 16

If $z=5+3i$ , show that $z\bar{z}=|z|^{2}$ .

Solution

Compact solution

Write the conjugate first, changing the sign of the imaginary part.
$\bar{z} = 5-3i$
Multiply $z$ by $\bar{z}$ . The two brackets are of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$z\bar{z} = (5+3i)(5-3i) = 5^{2}-(3i)^{2}$
Open the second square, squaring both the $3$ and the $i$ .
$(3i)^{2} = 9i^{2} = 9(-1) = -9$
Subtract. Taking away $-9$ is the same as adding $9$ .
$z\bar{z} = 25-(-9) = 25+9 = 34 \qquad \cdots(1)$
Now the other side. First find the modulus, using $|z|=\sqrt{x^{2}+y^{2}}$ with $x=5$ and $y=3$ .
$|z| = \sqrt{5^{2}+3^{2}} = \sqrt{34}$
Square it. Squaring a square root gives back the number inside, so the root and the square cancel.
$|z|^{2} = \left(\sqrt{34}\right)^{2} = 34 \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides give $34$ , which is what was to be shown.
$z\bar{z} = |z|^{2} = 34$

AnswerBoth sides give

34

, so the property holds.

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{5+3i} \\ &= 5-3i\end{aligned}

For the left side, multiply the conjugate pair.

\begin{aligned}z\bar{z} &= (5+3i)(5-3i) \\ &= 5^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 25-9i^2 \\ &= 25-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 25+9 \\ &= 34\end{aligned}

For the right side, use the modulus of the original number.

\begin{aligned}|z| &= \sqrt{5^2+3^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+9} \\ &= \sqrt{34}\end{aligned}

Square the modulus, as the right side requires.

\begin{aligned}|z|^2 &= \left(\sqrt{34}\right)^2 \\ &= 34\end{aligned}

The two sides agree.

z\bar{z}=|z|^2=34

Common mistake

Watch outThe property gives

|z|^{2}

, not

|z|

. Do not take a square root at the end.

Exercise 1.3

Question 1

Find the modulus of the following complex numbers:

(i)

4+3i

Solution

Compact solution

The modulus of $z=x+iy$ is $\sqrt{x^{2}+y^{2}}$ . Write down the two parts first, matching the number to that form.
$x = 4, \qquad y = 3$
Put them into the formula.
$|4+3i| = \sqrt{4^{2}+3^{2}}$
Square each part.
$4^{2} = 16, \qquad 3^{2} = 9$
Add the two squares.
$16+9 = 25$
Take the square root. Since $5\times 5=25$ , the root is exact.
$|4+3i| = \sqrt{25} = 5$

Answer

5

Compact solution

The real and imaginary coefficients are $4$ and $3$ . Use the nonnegative square root.

\begin{aligned}\left|4+3i\right| &= \sqrt{4^2+3^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5\end{aligned}

(ii)

-5-4i

Solution

Compact solution

Write down the two parts. Both are negative here, but that will not matter, because squaring removes a minus sign.
$x = -5, \qquad y = -4$
Put them into the formula $|z|=\sqrt{x^{2}+y^{2}}$ .
$|{-5}-4i| = \sqrt{(-5)^{2}+(-4)^{2}}$
Square each part. Minus times minus gives plus, so both squares are positive.
$(-5)^{2} = 25, \qquad (-4)^{2} = 16$
Add the two squares.
$25+16 = 41$
Take the square root. Since $41$ is not a perfect square, the root must stay as it is.
$|{-5}-4i| = \sqrt{41}$

Answer

\sqrt{41}

Compact solution

The real and imaginary coefficients are $-5$ and $-4$ . Use the nonnegative square root.

\begin{aligned}\left|-5-4i\right| &= \sqrt{(-5)^2+(-4)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+16} \\ &= \sqrt{41}\end{aligned}

Common mistake

Watch outA modulus cannot be negative. Here the negative coordinates are squared:

(-5)^{2}=25

and

(-4)^{2}=16

(iii)

\frac{3}{5}-\frac{4}{5}i

Solution

Compact solution

Write down the two parts. They are fractions here, but the formula does not change.
$x = \frac{3}{5}, \qquad y = -\frac{4}{5}$
Put them into the formula.
$\left|\frac{3}{5}-\frac{4}{5}i\right| = \sqrt{\left(\frac{3}{5}\right)^{2}+\left(-\frac{4}{5}\right)^{2}}$
Square each fraction. To square a fraction, square the top and square the bottom separately. The minus sign disappears in the second one.
$\left(\frac{3}{5}\right)^{2} = \frac{9}{25}, \qquad \left(-\frac{4}{5}\right)^{2} = \frac{16}{25}$
Add them. Both fractions already have the same bottom number $25$ , so add the tops and keep that bottom.
$\frac{9}{25}+\frac{16}{25} = \frac{25}{25} = 1$
Take the square root of $1$ .
$\left|\frac{3}{5}-\frac{4}{5}i\right| = \sqrt{1} = 1$

Answer

1

Compact solution

The real and imaginary coefficients are $\dfrac{3}{5}$ and $-\dfrac{4}{5}$ . Use the nonnegative square root.

\begin{aligned}\left|\dfrac{3}{5}-\dfrac{4}{5}i\right| &= \sqrt{\left(\frac{3}{5}\right)^2+\left(-\frac{4}{5}\right)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{\dfrac{9}{25}+\dfrac{16}{25}} \\ &= \sqrt{\frac{9+16}{25}} \\ &= \sqrt{\frac{25}{25}} \\ &= \sqrt{1} \\ &= 1\end{aligned}

Observation

ObservationA complex number with modulus

1

lies on the circle of radius

1

centered at the origin.

(iv)

-\sqrt{2}-\sqrt{3}i

Solution

Compact solution

Write down the two parts. Both are surds, but a surd is just an ordinary real number, so nothing about the method changes.
$x = -\sqrt{2}, \qquad y = -\sqrt{3}$
Put them into the formula.
$\left|-\sqrt{2}-\sqrt{3}i\right| = \sqrt{\left(-\sqrt{2}\right)^{2}+\left(-\sqrt{3}\right)^{2}}$
Square each part. Squaring a square root gives back the number inside, and the minus signs disappear.
$\left(-\sqrt{2}\right)^{2} = 2, \qquad \left(-\sqrt{3}\right)^{2} = 3$
Add the two squares.
$2+3 = 5$
Take the square root. Since $5$ is not a perfect square, the root stays as it is.
$\left|-\sqrt{2}-\sqrt{3}i\right| = \sqrt{5}$

Answer

\sqrt{5}

Compact solution

The two coefficients are $-\sqrt2$ and $-\sqrt3$ .

\begin{aligned}\left|-\sqrt2-\sqrt3i\right| &= \sqrt{(-\sqrt2)^2+(-\sqrt3)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{2+3} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-\sqrt a)^2=a\quad(a\geq0)$}} \\ &= \sqrt5\end{aligned}

Question 2

If $z_{1}=2+7i$ and $z_{2}=4-3i$ , then verify that:

(i)

\overline{z_{1}+z_{2}}=\bar{z_{1}}+\bar{z_{2}}

Solution

Compact solution

To verify, work out each side on its own and show they agree. Never assume they are equal. The left side says add first, then take the conjugate. So add.
$z_{1}+z_{2} = (2+7i)+(4-3i)$
Add the real parts together and the imaginary parts together.
$z_{1}+z_{2} = (2+4)+(7-3)i = 6+4i$
Now take the conjugate of that sum, changing the sign of the imaginary part.
$\overline{z_{1}+z_{2}} = 6-4i \qquad \cdots(1)$
The right side says take the conjugates first. Write each one.
$\bar{z_{1}} = 2-7i, \qquad \bar{z_{2}} = 4+3i$
Now add them, part by part. Note that $-7+3=-4$ , because starting at $-7$ and going up $3$ reaches $-4$ .
$\bar{z_{1}}+\bar{z_{2}} = (2+4)+(-7+3)i = 6-4i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides give the same number.
$\overline{z_{1}+z_{2}} = \bar{z_{1}}+\bar{z_{2}} = 6-4i$

AnswerBoth sides give

6-4i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}z_1+z_2 &= (2+7i)+(4-3i) \\ &= 2+7i+4-3i \\ &= (2+4)+(7-3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 6+4i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{z_1+z_2} &= \overline{6+4i} \\ &= 6-4i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=2-7i \\ \bar z_2=4+3i\end{gathered}

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}\bar z_1+\bar z_2 &= (2-7i)+(4+3i) \\ &= 2-7i+4+3i \\ &= (2+4)+(-7+3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 6-4i\end{aligned}

Both sides give $6-4i$ , so the required equality holds for these values.

\overline{z_1+z_2}=\bar z_1+\bar z_2=6-4i

(ii)

\overline{z_{1}z_{2}}=\bar{z_{1}}\,\bar{z_{2}}

Solution

Compact solution

The left side says multiply first, then take the conjugate. Multiply the brackets out term by term.
$z_{1}z_{2} = (2+7i)(4-3i) = 8-6i+28i-21i^{2}$
Replace $i^{2}$ by $-1$ . Note that $-21(-1)=+21$ , because minus times minus gives plus.
$-21i^{2} = 21$
Collect the real numbers and the $i$ terms.
$z_{1}z_{2} = (8+21)+(-6+28)i = 29+22i$
Now take the conjugate of that product.
$\overline{z_{1}z_{2}} = 29-22i \qquad \cdots(1)$
The right side says take the conjugates first. Write each one.
$\bar{z_{1}} = 2-7i, \qquad \bar{z_{2}} = 4+3i$
Multiply them out term by term.
$\bar{z_{1}}\,\bar{z_{2}} = (2-7i)(4+3i) = 8+6i-28i-21i^{2}$
Replace $i^{2}$ by $-1$ again, so the last term becomes $+21$ .
$-21i^{2} = 21$
Collect. Note that $6-28=-22$ , because taking $28$ from $6$ carries you $22$ below $0$ .
$\bar{z_{1}}\,\bar{z_{2}} = (8+21)+(6-28)i = 29-22i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides agree.
$\overline{z_{1}z_{2}} = \bar{z_{1}}\,\bar{z_{2}} = 29-22i$

AnswerBoth sides give

29-22i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Multiply each term in the first bracket by each term in the second.

\begin{aligned}z_1z_2 &= (2+7i)(4-3i) \\ &= 2(4)+2(-3i)+(7i)(4)+(7i)(-3i) \\ &= 8-6i+28i-21i^2 \\ &= 8-6i+28i-21(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 8-6i+28i+21 \\ &= (8+21)+(-6+28)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 29+22i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{z_1z_2} &= \overline{29+22i} \\ &= 29-22i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=2-7i \\ \bar z_2=4+3i\end{gathered}

Multiply each term in the first bracket by each term in the second.

\begin{aligned}\bar z_1\bar z_2 &= (2-7i)(4+3i) \\ &= 2(4)+2(3i)+(-7i)(4)+(-7i)(3i) \\ &= 8+6i-28i-21i^2 \\ &= 8+6i-28i-21(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 8+6i-28i+21 \\ &= (8+21)+(6-28)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 29-22i\end{aligned}

Both sides give $29-22i$ , so the required equality holds for these values.

\overline{z_1z_2}=\bar z_1\bar z_2=29-22i

(iii)

\overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\bar{z_{1}}}{\bar{z_{2}}}

Solution

Compact solution

The left side says divide first, then take the conjugate. Begin with the division.
$\frac{z_{1}}{z_{2}} = \frac{2+7i}{4-3i}$
There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate of the bottom, which is $4+3i$ .
$\frac{2+7i}{4-3i} = \frac{2+7i}{4-3i}\times\frac{4+3i}{4+3i}$
Work out the numerator, multiplying term by term.
$(2+7i)(4+3i) = 8+6i+28i+21i^{2}$
Replace $i^{2}$ by $-1$ , so the last term is $-21$ .
$21i^{2} = -21$
Collect the numerator. Note that $8-21=-13$ .
$(2+7i)(4+3i) = (8-21)+(6+28)i = -13+34i$
Now the denominator, using $(a-b)(a+b)=a^{2}-b^{2}$ .
$(4-3i)(4+3i) = 4^{2}-(3i)^{2} = 16+9 = 25$
Put the numerator over the denominator and split into two parts.
$\frac{z_{1}}{z_{2}} = \frac{-13+34i}{25} = -\frac{13}{25}+\frac{34}{25}i$
Now take the conjugate of that, changing the sign of the imaginary part only.
$\overline{\left(\frac{z_{1}}{z_{2}}\right)} = -\frac{13}{25}-\frac{34}{25}i \qquad \cdots(1)$
The right side says take the conjugates first. Write each one.
$\bar{z_{1}} = 2-7i, \qquad \bar{z_{2}} = 4+3i$
Divide them. The conjugate of the new bottom $4+3i$ is $4-3i$ , so multiply top and bottom by that.
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{2-7i}{4+3i}\times\frac{4-3i}{4-3i}$
Work out the numerator.
$(2-7i)(4-3i) = 8-6i-28i+21i^{2}$
Replace $i^{2}$ by $-1$ and collect. Note that $-6-28=-34$ .
$(2-7i)(4-3i) = (8-21)+(-6-28)i = -13-34i$
The denominator is the same as before.
$(4+3i)(4-3i) = 16+9 = 25$
So the right side comes to this.
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{-13-34i}{25} = -\frac{13}{25}-\frac{34}{25}i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides agree, and this is the value they share.
$\overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{\bar{z_{1}}}{\bar{z_{2}}} = -\frac{13}{25}-\frac{34}{25}i$

AnswerBoth sides give

-\dfrac{13}{25}-\dfrac{34}{25}i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Multiply numerator and denominator by the conjugate $4+3i$ .

\begin{aligned}\frac{z_1}{z_2} &= \frac{2+7i}{4-3i} \\ &= \frac{2+7i}{4-3i}\cdot\frac{4+3i}{4+3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{4+3i}{4+3i}=1$}} \\ &= \frac{(2+7i)(4+3i)}{(4-3i)(4+3i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(2+7i)(4+3i) &= 2(4)+2(3i)+(7i)(4)+(7i)(3i) \\ &= 8+6i+28i+21i^2 \\ &= 8+6i+28i+21(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 8+6i+28i-21 \\ &= (8-21)+(6+28)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -13+34i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(4-3i)(4+3i) &= 4^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 16-9i^2 \\ &= 16-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 16+9 \\ &= 25\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{z_1}{z_2} &= \frac{-13+34i}{25} \\ &= -\frac{13}{25}+\frac{34}{25}i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{\frac{z_1}{z_2}} &= \overline{-\dfrac{13}{25}+\dfrac{34}{25}i} \\ &= -\dfrac{13}{25}-\dfrac{34}{25}i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=2-7i \\ \bar z_2=4+3i\end{gathered}

Multiply numerator and denominator by the conjugate $4-3i$ .

\begin{aligned}\frac{\bar z_1}{\bar z_2} &= \frac{2-7i}{4+3i} \\ &= \frac{2-7i}{4+3i}\cdot\frac{4-3i}{4-3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{4-3i}{4-3i}=1$}} \\ &= \frac{(2-7i)(4-3i)}{(4+3i)(4-3i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(2-7i)(4-3i) &= 2(4)+2(-3i)+(-7i)(4)+(-7i)(-3i) \\ &= 8-6i-28i+21i^2 \\ &= 8-6i-28i+21(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 8-6i-28i-21 \\ &= (8-21)+(-6-28)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -13-34i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(4+3i)(4-3i) &= 4^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 16-9i^2 \\ &= 16-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 16+9 \\ &= 25\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{\bar z_1}{\bar z_2} &= \frac{-13-34i}{25} \\ &= -\frac{13}{25}-\frac{34}{25}i\end{aligned}

Both sides give $-\dfrac{13}{25}-\dfrac{34}{25}i$ , so the required equality holds for these values.

\overline{\frac{z_1}{z_2}}=\frac{\bar z_1}{\bar z_2}=-\dfrac{13}{25}-\dfrac{34}{25}i

Question 3

If $z=5-2i$ , then verify that:

(i)

\bar{\bar{z}}=z

Solution

Compact solution

Take the conjugate once. Keep the real part and change the sign of the imaginary part.
$\bar{z} = 5+2i$
Now take the conjugate again, of the number just found. The real part $5$ stays, and the imaginary part $+2$ changes sign to $-2$ .
$\bar{\bar{z}} = \overline{5+2i} = 5-2i$
That is the number we started with, so the property holds.
$\bar{\bar{z}} = 5-2i = z$

Answer

\bar{\bar{z}}=5-2i=z

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{5-2i} \\ &= 5+2i\end{aligned}

Conjugate once more; the imaginary sign changes back.

\begin{aligned}\overline{\bar{z}} &= \overline{5+2i} \\ &= 5-2i \\ &= z\end{aligned}

(ii)

|z|=|\bar{z}|

Solution

Compact solution

Use $|z|=\sqrt{x^{2}+y^{2}}$ with $x=5$ and $y=-2$ .
$|z| = \sqrt{5^{2}+(-2)^{2}}$
Square each part. Minus times minus gives plus, so the second square is positive.
$5^{2} = 25, \qquad (-2)^{2} = 4$
Add and take the root. Since $29$ is not a perfect square, the root stays.
$|z| = \sqrt{29} \qquad \cdots(1)$
Now write the conjugate and take its modulus. Here $x=5$ and $y=+2$ .
$\bar{z} = 5+2i, \qquad |\bar{z}| = \sqrt{5^{2}+2^{2}}$
Square each part and add.
$|\bar{z}| = \sqrt{25+4} = \sqrt{29} \qquad \cdots(2)$
From $(1)$ and $(2)$ both are the same. The squaring removed the only difference between them, which was a sign.
$|z| = |\bar{z}| = \sqrt{29}$

AnswerBoth sides give

\sqrt{29}

, so the property holds.

Compact solution

The real and imaginary coefficients are $5$ and $-2$ . Use the nonnegative square root.

\begin{aligned}|z| &= \sqrt{5^2+(-2)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{5-2i} \\ &= 5+2i\end{aligned}

The real and imaginary coefficients are $5$ and $2$ . Use the nonnegative square root.

\begin{aligned}\left|\bar{z}\right| &= \sqrt{5^2+2^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

The two distances are equal.

|z|=\left|\bar{z}\right|=\sqrt{29}

(iii)

|z|=|{-z}|

Solution

Compact solution

The modulus of $z$ was found in part (ii).
$|z| = \sqrt{29} \qquad \cdots(1)$
Now write $-z$ . A minus in front of the whole number changes the sign of both parts.
$-z = -(5-2i) = -5+2i$
Take its modulus, with $x=-5$ and $y=2$ .
$|{-z}| = \sqrt{(-5)^{2}+2^{2}}$
Square each part. The minus sign disappears in the first square.
$(-5)^{2} = 25, \qquad 2^{2} = 4$
Add and take the root.
$|{-z}| = \sqrt{29} \qquad \cdots(2)$
From $(1)$ and $(2)$ both are the same. Turning a point through $180^\circ$ about the origin does not change how far it is from the origin.
$|z| = |{-z}| = \sqrt{29}$

AnswerBoth sides give

\sqrt{29}

, so the property holds.

Compact solution

The real and imaginary coefficients are $5$ and $-2$ . Use the nonnegative square root.

\begin{aligned}|z| &= \sqrt{5^2+(-2)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(5-2i) \\ &= -5+2i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a-b)=-a+b$}}\end{aligned}

The real and imaginary coefficients are $-5$ and $2$ . Use the nonnegative square root.

\begin{aligned}\left|-z\right| &= \sqrt{(-5)^2+2^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

The two distances are equal.

|z|=\left|-z\right|=\sqrt{29}

(iv)

z\bar{z}=|z|^{2}

Solution

Compact solution

Write the conjugate first.
$\bar{z} = 5+2i$
Multiply $z$ by $\bar{z}$ . The two brackets are of the form $(a-b)(a+b)$ , which is always $a^{2}-b^{2}$ .
$z\bar{z} = (5-2i)(5+2i) = 5^{2}-(2i)^{2}$
Open the second square, squaring both the $2$ and the $i$ .
$(2i)^{2} = 4i^{2} = 4(-1) = -4$
Subtract. Taking away $-4$ is the same as adding $4$ , so the answer is an ordinary real number.
$z\bar{z} = 25-(-4) = 25+4 = 29 \qquad \cdots(1)$
Now the other side. The modulus was found in part (ii).
$|z| = \sqrt{29}$
Square it. Squaring a square root gives back the number inside, so the two cancel.
$|z|^{2} = \left(\sqrt{29}\right)^{2} = 29 \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides give the same number.
$z\bar{z} = |z|^{2} = 29$

AnswerBoth sides give

29

, so the property holds.

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{5-2i} \\ &= 5+2i\end{aligned}

For the left side, multiply the conjugate pair.

\begin{aligned}z\bar{z} &= (5-2i)(5+2i) \\ &= 5^2-(2i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 25-4i^2 \\ &= 25-4(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 25+4 \\ &= 29\end{aligned}

For the right side, use the modulus of the original number.

\begin{aligned}|z| &= \sqrt{5^2+(-2)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

Square the modulus, as the right side requires.

\begin{aligned}|z|^2 &= \left(\sqrt{29}\right)^2 \\ &= 29\end{aligned}

The two sides agree.

z\bar{z}=|z|^2=29

Common mistake

Watch outThe right side is

|z|^{2}

, not

|z|

. Do not take a square root at the end.

(v)

|z|=|{-\bar{z}}|

Solution

Compact solution

The modulus of $z$ was found in part (ii).
$|z| = \sqrt{29} \qquad \cdots(1)$
Now build $-\bar{z}$ in two moves. First take the conjugate, which changes the sign of the imaginary part only.
$\bar{z} = 5+2i$
Then put a minus in front of that, which changes the sign of both parts.
$-\bar{z} = -(5+2i) = -5-2i$
Take its modulus, with $x=-5$ and $y=-2$ .
$|{-\bar{z}}| = \sqrt{(-5)^{2}+(-2)^{2}}$
Square each part. Both minus signs disappear.
$(-5)^{2} = 25, \qquad (-2)^{2} = 4$
Add and take the root.
$|{-\bar{z}}| = \sqrt{29} \qquad \cdots(2)$
From $(1)$ and $(2)$ both are the same. Whatever combination of sign changes you make, the two numbers being squared are unchanged.
$|z| = |{-\bar{z}}| = \sqrt{29}$

AnswerBoth sides give

\sqrt{29}

, so the property holds.

Compact solution

The real and imaginary coefficients are $5$ and $-2$ . Use the nonnegative square root.

\begin{aligned}|z| &= \sqrt{5^2+(-2)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z} &= \overline{5-2i} \\ &= 5+2i\end{aligned}

Negate the conjugate by changing both signs.

\begin{aligned}-\bar{z} &= -(5+2i) \\ &= -5-2i\end{aligned}

The real and imaginary coefficients are $-5$ and $-2$ . Use the nonnegative square root.

\begin{aligned}\left|-\bar{z}\right| &= \sqrt{(-5)^2+(-2)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+4} \\ &= \sqrt{29}\end{aligned}

The two distances are equal.

|z|=\left|-\bar{z}\right|=\sqrt{29}

Question 4

If $z=4-3i$ , then verify that $|z|=|{-z}|=\left|\bar{\bar{z}}\right|=|{-\bar{z}}|$ .

Solution

Compact solution

There are four numbers to compare, so build each one first before finding any modulus. Start with $z$ itself.
$z = 4-3i$
Change both signs to get $-z$ .
$-z = -(4-3i) = -4+3i$
Take the conjugate twice. Taking it once gives $4+3i$ , and taking it again changes the sign back.
$\bar{z} = 4+3i, \qquad \bar{\bar{z}} = 4-3i$
Now put a minus in front of the conjugate, which changes both signs of $4+3i$ .
$-\bar{z} = -(4+3i) = -4-3i$
Now find the four moduli, using $|z|=\sqrt{x^{2}+y^{2}}$ each time. Take $z$ first, with $x=4$ and $y=-3$ .
$|z| = \sqrt{4^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$
Now $-z$ , with $x=-4$ and $y=3$ . Squaring removes both signs.
$|{-z}| = \sqrt{(-4)^{2}+3^{2}} = \sqrt{16+9} = 5$
Now $\bar{\bar{z}}$ , which turned out to be $z$ itself, so its modulus must be the same.
$\left|\bar{\bar{z}}\right| = \sqrt{4^{2}+(-3)^{2}} = 5$
Finally $-\bar{z}$ , with $x=-4$ and $y=-3$ .
$|{-\bar{z}}| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = 5$
All four give the same answer, which is what was to be verified.
$|z| = |{-z}| = \left|\bar{\bar{z}}\right| = |{-\bar{z}}| = 5$

AnswerAll four give

5

, so the chain holds.

Compact solution

First form the numbers whose moduli are required.

\begin{gathered}z=4-3i \\ -z=-4+3i \\ \bar z=4+3i \\ \bar{\bar z}=4-3i \\ -\bar z=-4-3i\end{gathered}

The real and imaginary coefficients are $4$ and $-3$ . Use the nonnegative square root.

\begin{aligned}|z| &= \sqrt{4^2+(-3)^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5\end{aligned}

The other numbers differ only in signs. Squaring each coefficient removes those sign changes.

\begin{gathered}|-z|=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=5 \\ |\bar{\bar z}|=\sqrt{4^2+(-3)^2}=5 \\ |-\bar z|=\sqrt{(-4)^2+(-3)^2}=5\end{gathered}

Hence all four moduli are equal.

|z|=|-z|=|\bar{\bar z}|=|-\bar z|=5

Common mistake

Watch outDo not shorten this by claiming the four numbers are equal. They are not. Only their moduli are equal, because each sign change disappears when the parts are squared.

Question 5

If $z_{1}=2+3i$ and $z_{2}=-1+i$ , then evaluate:

(i)

\operatorname{Re}(z_{1}z_{2})

Solution

Compact solution

The real part cannot be read off until the product is worked out, so multiply first. Take the brackets out term by term.
$z_{1}z_{2} = (2+3i)(-1+i) = 2(-1)+2(i)+3i(-1)+3i(i)$
Work out the first three products. Plus times minus gives minus in the first and third.
$2(-1) = -2, \quad 2(i) = 2i, \quad 3i(-1) = -3i$
Now the fourth. The two $i$ terms multiply to give $i^{2}$ .
$3i(i) = 3i^{2}$
Put the four products together.
$z_{1}z_{2} = -2+2i-3i+3i^{2}$
Replace $i^{2}$ by $-1$ , so the last term becomes $-3$ .
$3i^{2} = 3(-1) = -3$
Collect the real numbers and the $i$ terms.
$z_{1}z_{2} = (-2-3)+(2-3)i = -5-i$
The real part is the piece with no $i$ attached to it.
$\operatorname{Re}(z_{1}z_{2}) = -5$

Answer

-5

Compact solution

Multiply each term in the first bracket by each term in the second.

\begin{aligned}z_1z_2 &= (2+3i)(-1+i) \\ &= 2(-1)+2(i)+(3i)(-1)+(3i)(i) \\ &= -2+2i-3i+3i^2 \\ &= -2+2i-3i+3(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -2+2i-3i-3 \\ &= (-2-3)+(2-3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -5-i\end{aligned}

The real coefficient is $-5$ .

\operatorname{Re}(z_1z_2)=-5

(ii)

\operatorname{Im}(z_{1}z_{2})

Solution

Compact solution

This is the same product as in part (i), so the work is already done.
$z_{1}z_{2} = -5-i$
The imaginary part is the number multiplying $i$ , without the $i$ itself. Here the term is $-i$ , which means $-1$ multiplied by $i$ .
$-i = (-1)i$
So the imaginary part is $-1$ .
$\operatorname{Im}(z_{1}z_{2}) = -1$

Answer

-1

Compact solution

Multiply each term in the first bracket by each term in the second.

\begin{aligned}z_1z_2 &= (2+3i)(-1+i) \\ &= 2(-1)+2(i)+(3i)(-1)+(3i)(i) \\ &= -2+2i-3i+3i^2 \\ &= -2+2i-3i+3(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -2+2i-3i-3 \\ &= (-2-3)+(2-3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -5-i\end{aligned}

Since $-i=(-1)i$ , the imaginary coefficient is $-1$ , not $-i$ .

\operatorname{Im}(z_1z_2)=-1

Common mistake

Watch outThe answer is

-1

, not

-i

. The imaginary part is a real number, and the

i

is never part of it.

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Class 10 Mathematics

Explore the book

Complex Numbers

Quadratic Equations and Inequalities

Matrices and Determinants

Functions and Graphs

Algebraic Fractions

Vectors in Plane

Trigonometry

Chords and Arcs of a Circle

Tangent and Angles of a Circle

Practical Geometry of Circles

Information Handling

Probability

The complex plane

Conjugate of a complex number

Properties of the conjugate

Modulus of a complex number

Properties of the modulus

Why conjugation preserves sums

Why conjugation preserves products

Why conjugation preserves quotients

Why conjugating twice returns the number

Why the modulus is a distance

Why ∣z∣|z|, ∣zˉ∣|\bar{z}|, ∣−z∣|{-z}| and ∣−zˉ∣|{-\bar{z}}| are all equal

Why zzˉ=∣z∣2z\bar{z}=|z|^{2}

The Argand diagram

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Question 1

Question 2

Question 3

Question 4

Question 5

Class 10 Mathematics

Explore the book

Complex Numbers

Quadratic Equations and Inequalities

Matrices and Determinants

Functions and Graphs

Algebraic Fractions

Vectors in Plane

Trigonometry

Chords and Arcs of a Circle

Tangent and Angles of a Circle

Practical Geometry of Circles

Information Handling

Probability

Class 10
Mathematics

Why $|z|$ , $|\bar{z}|$ , $|{-z}|$ and $|{-\bar{z}}|$ are all equal

Why $z\bar{z}=|z|^{2}$

Class 10
Mathematics