Class 10 Exercise 1.1 Solutions — Complex Numbers

Concepts

Definition

The imaginary unit

The imaginary unit is denoted by $i$ and satisfies $i^{2}=-1$ By convention, the principal square root $\sqrt{-1}$ is $i$ . The equation $t^{2}=-1$ has two solutions: $t=i$ and $t=-i$ .

No real number has a square equal to $-1$ , so $i$ is a new kind of number. It is not on the real number line.

In many classrooms in Pakistan $i$ is read as iota. Its usual name in books around the world is the imaginary unit. Both names mean the same thing.

Examples

$i\times i=i^{2}=-1$ .
$3i\times 3i=9i^{2}=9(-1)=-9$ .
$\sqrt{-9}=i\sqrt{9}=3i$ .

Definition

Complex numbers and rectangular form

A complex number is any number that can be written as $z=x+iy$ where $x$ and $y$ are real numbers and $i$ is the imaginary unit. This way of writing it is called the rectangular form.

Here $x$ is called the real part of $z$ , and $y$ is called the imaginary part of $z$ . We write $\operatorname{Re}(z)=x, \qquad \operatorname{Im}(z)=y$

Be careful. The imaginary part is $y$ only. The $i$ is not part of it. So for $z=5-7i$ we have $\operatorname{Im}(z)=-7$ , not $-7i$ .

The form $x+yi$ means the same as $x+iy$ . Both are correct.

Examples

For $z=3+4i$ : $\ \operatorname{Re}(z)=3$ and $\operatorname{Im}(z)=4$ .
For $z=5-7i$ : $\ \operatorname{Re}(z)=5$ and $\operatorname{Im}(z)=-7$ .
For $z=-2i$ : write it as $0+(-2)i$ , so $\operatorname{Re}(z)=0$ and $\operatorname{Im}(z)=-2$ .
For $z=6$ : write it as $6+0i$ , so $\operatorname{Re}(z)=6$ and $\operatorname{Im}(z)=0$ .

Definition

Real and purely imaginary complex numbers

Let $z=x+iy$ , where $x$ and $y$ are real. If $y=0$ , then $z$ is a real number. If $x=0$ and $y\neq 0$ , then $z$ is called purely imaginary.

So every real number is also a complex number. For example $7=7+0i$ .

We ask for $y\neq 0$ in the second case so that $0$ is not counted as purely imaginary. The number $0$ is real.

Examples

$z=9+0i$ is real, because its imaginary part is $0$ .
$z=0+4i$ is purely imaginary, because its real part is $0$ and its imaginary part is not $0$ .
$z=2+3i$ is neither real nor purely imaginary, because both its real and imaginary parts are nonzero.

Definition

Equality of complex numbers

Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ , with all four coefficients real. Then $z_{1}=z_{2}\iff x_{1}=x_{2}\ \text{and}\ y_{1}=y_{2}$ Thus, two complex numbers are equal if and only if their corresponding real and imaginary parts are equal.

This is the rule that lets you find two unknowns. One equation between complex numbers gives you two real equations.

The rule only works when all four numbers $x_{1}$ , $y_{1}$ , $x_{2}$ and $y_{2}$ are real. Always check this first.

Examples

If $\alpha+i\beta=-2+5i$ , where $\alpha$ and $\beta$ are real, then $\alpha=-2$ and $\beta=5$ .
If $x+iy=7$ , where $x$ and $y$ are real, then $x=7$ and $y=0$ .
$3+4i\neq 4+3i$ , because the real parts $3$ and $4$ are not equal.

Deep understanding

Why we needed a new kind of number

Look at $x^{2}-1=0$ . Add $1$ to both sides and you get $x^{2}=1$ . Two numbers work here. First $x=1$ , because $1\times 1=1$ . Second $x=-1$ , because $(-1)\times(-1)=1$ . So this equation has two real answers.

Now change one sign. Look at $x^{2}+1=0$ . Take $1$ from both sides and you get $x^{2}=-1$ .

Which real number works? A positive number times itself is positive, like $3\times 3=9$ . A negative number times itself is also positive, like $(-3)\times(-3)=9$ , because minus times minus gives plus. And $0\times 0=0$ . So the square of a real number is never negative. No real number works.

You can also see this on a graph. The curve $f(x)=x^{2}-1$ cuts the $x$ -axis at $(-1,0)$ and $(1,0)$ , so that equation has two real answers. The curve $f(x)=x^{2}+1$ never touches the $x$ -axis, so that equation has no real answer.

So mathematicians made a new number for this job. They called it $i$ , and gave it one rule: $i^{2}=-1$ . This defining relation lets us extend familiar algebra to complex numbers.

Powers of $i$

Integer powers of $i$ follow from $i^{2}=-1$ . For example, the first four positive powers are $i$ , $-1$ , $-i$ , and $1$ .

The first power of any number is the number itself, so $i^{1}=i$ . The second power is the rule itself, so $i^{2}=-1$ .

For the third power, take out one factor of $i$ and use the rule on what is left. $\begin{aligned}i^{3} &= i^{2}\cdot i \\ &= (-1)\cdot i \\ &= -i\end{aligned}$

For the fourth power, write it as the square of $i^{2}$ . $\begin{aligned}i^{4} &= \left(i^{2}\right)^{2} \\ &= (-1)^{2} \\ &= 1\end{aligned}$

For a nonnegative integer exponent, an even power has the form $2k$ , while an odd power has the form $2k+1$ , with $k$ a nonnegative integer. Write $i^{2k}=(i^{2})^{k}$ ; in an odd power, leave one factor of $i$ outside.

You are then left with $(-1)^{k}$ , which is easy to read off. It is $1$ when $k$ is even and $-1$ when $k$ is odd, because minus signs cancel in pairs and an odd one is always left over.

Rule. For

k=0,1,2,\ldots

i^{2k}=(-1)^{k},\qquad i^{2k+1}=(-1)^{k}i

Take

i^{23}

. It is odd, and

23=2(11)+1

, so

i^{23}=\left(i^{2}\right)^{11}\cdot i=(-1)^{11}\cdot i=-i

, because

11

is odd.

Negative powers and the reciprocal of $i$

For a nonzero number $a$ and a positive integer $n$ , a negative exponent means a reciprocal: $a^{-n}=\frac{1}{a^{n}}$ In particular, $i^{-n}=1/i^{n}$ because $i\neq0$ .

After simplifying the denominator, we may obtain $1/i$ or $1/(-i)$ . These are valid exact expressions. To write them in rectangular form, replace the imaginary denominator with a real one.

To clear it, multiply the top and the bottom by $i$ . You may do this because $\frac{i}{i}=1$ , and multiplying by $1$ changes how a number looks but not its value. $\begin{aligned}\frac{1}{i} &= \frac{1}{i}\times\frac{i}{i} \\ &= \frac{1\times i}{i\times i} \\ &= \frac{i}{i^{2}} \\ &= \frac{i}{-1} \\ &= -i\end{aligned}$

Read the last line slowly. Dividing by $-1$ is the same as multiplying by $-1$ . It just changes the sign.

Square roots of negative numbers

Let $a$ be a positive number. To write $\sqrt{-a}$ using $i$ , first pull the minus sign out as a factor of $-1$ . $\begin{aligned}\sqrt{-a} &= \sqrt{a\times(-1)} \\ &= \sqrt{a}\times\sqrt{-1} \\ &= i\sqrt{a}\end{aligned}$

Here $a>0$ , and we use the specific principal-root identity $\sqrt{-a}=\sqrt{a}\,\sqrt{-1}=i\sqrt{a}$ . Do not extend the real product rule to two negative factors: $\sqrt{-1}\sqrt{-1}=-1$ , whereas $\sqrt{(-1)(-1)}=1$ .

For example $\sqrt{-4}=\sqrt{4}\times\sqrt{-1}=2i$ , because $\sqrt{4}=2$ . And $\sqrt{-7}=i\sqrt{7}$ . Here $\sqrt{7}$ must stay as it is, because $7$ is not a perfect square.

Both $i\sqrt{7}$ and $\sqrt{7}\,i$ are correct. We prefer $i\sqrt{7}$ in our working to make it clear that $i$ is outside the radical. Neither expression means $\sqrt{7i}$ .

From the number line to the complex plane

A real number can be represented on a line. A complex number needs two real coordinates: $x$ gives its horizontal position, and $y$ gives its vertical position.

The geometric interpretation was developed independently by several mathematicians. Wessel published a plane interpretation in 1799; Argand developed his independent account in 1806. Their work helps explain why complex numbers are represented as points, not as positions on the real number line.

Examples

Example 1

Simplify the following:

(i)

i^{7}

Solution

Compact solution

The exponent $7$ is odd. Take one factor of $i$ out, so that an even power is left behind. An even power can always be written using $i^{2}$ , and $i^{2}=-1$ is the only rule we need.
$i^{7} = i^{6}\cdot i$
Halve the even exponent, because $6=2\times 3$ . This is the rule $a^{mn}=\left(a^{m}\right)^{n}$ .
$i^{7} = \left(i^{2}\right)^{3}\cdot i$
Put $-1$ in place of $i^{2}$ , which is the definition of $i$ .
$i^{7} = (-1)^{3}\cdot i$
$3$ copies of $-1$ multiplied together. Minus signs cancel in pairs, and $3$ is odd, so one minus is left over.
$(-1)^{3} = -1$
Multiplying by $-1$ just changes the sign.
$i^{7} = -i$

Answer

-i

Compact solution

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{7} &= i^{6}\cdot i \\ &= (i^2)^{3}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{3}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (-1)\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}} \\ &= -i\end{aligned}

(ii)

i^{8}

Solution

Compact solution

The exponent $8$ is even, so the whole power can be written using $i^{2}$ . Halve the exponent, because $8=2\times 4$ .
$i^{8} = \left(i^{2}\right)^{4}$
Put $-1$ in place of $i^{2}$ .
$i^{8} = (-1)^{4}$
$4$ copies of $-1$ cancel in pairs, and $4$ is even, so none is left over.
$i^{8} = 1$

Answer

1

Compact solution

The exponent is even, so write the power in terms of $i^2$ .

\begin{aligned}i^{8} &= (i^2)^{4} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{4} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}}\end{aligned}

(iii)

i^{17}

Solution

Compact solution

The exponent $17$ is odd. Take one factor of $i$ out, leaving the even power $16$ behind.
$i^{17} = i^{16}\cdot i$
Halve the even exponent, because $16=2\times 8$ .
$i^{17} = \left(i^{2}\right)^{8}\cdot i$
Put $-1$ in place of $i^{2}$ .
$i^{17} = (-1)^{8}\cdot i$
$8$ copies of $-1$ cancel in pairs, and $8$ is even, so the result is $1$ .
$i^{17} = 1\cdot i = i$

Answer

i

Compact solution

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{17} &= i^{16}\cdot i \\ &= (i^2)^{8}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{8}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

(iv)

i^{-25}

Solution

Compact solution

The power is negative, so first turn the number upside down. This is the rule $a^{-n}=\frac{1}{a^{n}}$ . Now the power at the bottom is positive.
$i^{-25} = \frac{1}{i^{25}}$
Work on the bottom. The exponent $25$ is odd, so take one factor of $i$ out, leaving the even power $24$ .
$i^{25} = i^{24}\cdot i$
Halve the even exponent, because $24=2\times 12$ .
$i^{25} = \left(i^{2}\right)^{12}\cdot i$
Put $-1$ in place of $i^{2}$ . $12$ minus signs cancel in pairs, so the result is $1$ .
$i^{25} = (-1)^{12}\cdot i = i$
Substitute this value into the original expression.
$i^{-25} = \frac{1}{i}$
This exact fraction is not yet in rectangular form. Clear it by multiplying top and bottom by $i$ . You may do this because $\frac{i}{i}=1$ . Then use $i^{2}=-1$ , and dividing by $-1$ changes the sign.
$\begin{aligned}i^{-25} &= \frac{1}{i}\cdot\frac{i}{i} \\ &= \frac{i}{i^{2}} \\ &= \frac{i}{-1} \\ &= -i\end{aligned}$

Answer

-i

Compact solution

First rewrite the negative power as a reciprocal.

i^{-25}=\frac{1}{i^{25}}

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{25} &= i^{24}\cdot i \\ &= (i^2)^{12}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{12}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

Now substitute into the reciprocal and simplify.

\begin{aligned}i^{-25} &= \frac{1}{i} \\ &= \frac{1}{i}\cdot\frac{i}{i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{i}{i}=1$}} \\ &= \frac{i}{i^2} \\ &= \frac{i}{(-1)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -i\end{aligned}

Example 2

If $(2x-1)+(y+4)i=5+7i$ , find the values of $x$ and $y$ . Assume $x$ and $y$ are real.

Solution

Compact solution

Both sides are already written as a real number plus a real number times $i$ , so we can compare them at once. The real part is the piece with no $i$ on it.
$\text{Left real part} = 2x-1, \qquad \text{Right real part} = 5$
The imaginary part is the number multiplying $i$ , without the $i$ itself.
$\text{Left imaginary part} = y+4, \qquad \text{Right imaginary part} = 7$
Two complex numbers are equal only when both parts match. So one complex equation gives two real equations.
$2x-1=5 \qquad \text{and} \qquad y+4=7$
Take the first equation. To leave $2x$ alone, add $1$ to both sides. What you do to one side you must do to the other.
$2x-1+1 = 5+1$
On the left $-1+1=0$ , so only $2x$ is left. On the right $5+1=6$ .
$2x = 6$
To leave $x$ alone, divide both sides by $2$ .
$\frac{2x}{2} = \frac{6}{2}$
On the left the two $2$ s cancel. On the right $6\div 2=3$ .
$x = 3$
Now the second equation. Take $4$ from both sides.
$y+4-4 = 7-4$
On the left $4-4=0$ , so $y$ is left. On the right $7-4=3$ .
$y = 3$

Answer

x=3

and

y=3

Compact solution

Since $x,y\in\mathbb R$ , compare the real parts and solve for $x$ .

\begin{aligned}2x-1&=5 \\ 2x-1+1&=5+1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add }1\text{ on both sides}$}} \\ 2x&=6 \\ \frac{2x}{2}&=\frac{6}{2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by }2$}} \\ x&=3\end{aligned}

Now compare the imaginary coefficients and solve for $y$ .

\begin{aligned}y+4&=7 \\ y+4-4&=7-4 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{subtract }4\text{ on both sides}$}} \\ y&=3\end{aligned}

Check

Check the answer

\begin{aligned}(2(3)-1)+((3)+4)i &= (6-1)+(7)i \\ &= 5+7i \quad \checkmark\end{aligned}

Put

x=3

and

y=3

back into the left side. It gives the right side exactly, so both answers are correct.

Exercise 1.1

Question 1

Simplify the following:

(i)

i^{5}

Solution

Compact solution

The exponent $5$ is odd. Take one factor of $i$ out, so that an even power is left behind. An even power can always be written using $i^{2}$ , and $i^{2}=-1$ is the only rule we need.
$i^{5} = i^{4}\cdot i$
Halve the even exponent, because $4=2\times 2$ . This is the rule $a^{mn}=\left(a^{m}\right)^{n}$ .
$i^{5} = \left(i^{2}\right)^{2}\cdot i$
Put $-1$ in place of $i^{2}$ , which is the definition of $i$ .
$i^{5} = (-1)^{2}\cdot i$
$2$ copies of $-1$ multiplied together. Minus times minus gives plus, so they cancel.
$i^{5} = 1\cdot i$
Multiplying by $1$ changes nothing.
$i^{5} = i$

Answer

i

Compact solution

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{5} &= i^{4}\cdot i \\ &= (i^2)^{2}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{2}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

Video solution

Open this video on YouTube if it does not play here.

(ii)

i^{16}

Solution

Compact solution

The exponent $16$ is even, so the whole power can be written using $i^{2}$ . Halve the exponent, because $16=2\times 8$ .
$i^{16} = \left(i^{2}\right)^{8}$
Put $-1$ in place of $i^{2}$ .
$i^{16} = (-1)^{8}$
$8$ copies of $-1$ cancel in pairs, and $8$ is even, so none is left over.
$i^{16} = 1$

Answer

1

Compact solution

The exponent is even, so write the power in terms of $i^2$ .

\begin{aligned}i^{16} &= (i^2)^{8} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{8} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}}\end{aligned}

(iii)

(-i)^{-19}

Solution

Compact solution

Keep track of three features: a minus sign inside the bracket, a negative power, and a large power. Deal with the negative power first. It means turn the bracket upside down.
$(-i)^{-19} = \frac{1}{(-i)^{19}}$
Now work on the bottom. A minus sign in front of something means that thing times $-1$ , so the bracket is really two factors multiplied.
$-i = (-1)\cdot i$
When a product is raised to a power, each factor can be raised to that power on its own. This is the rule $(ab)^{n}=a^{n}b^{n}$ .
$(-i)^{19} = (-1)^{19}\cdot i^{19}$
Take the first factor. $19$ copies of $-1$ cancel in pairs, and $19$ is odd, so one minus is left over.
$(-1)^{19} = -1$
Now the second factor. The exponent $19$ is odd, so take one factor of $i$ out, leaving the even power $18$ .
$i^{19} = i^{18}\cdot i$
Halve the even exponent, because $18=2\times 9$ , then put $-1$ in place of $i^{2}$ . Nine is odd, so one minus is left over.
$i^{19} = \left(i^{2}\right)^{9}\cdot i = (-1)^{9}\cdot i = -i$
Multiply the two factors. Minus times minus gives plus.
$(-i)^{19} = (-1)\cdot(-i) = i$
Put that back at the bottom.
$(-i)^{-19} = \frac{1}{i}$
This exact fraction is not yet in rectangular form. Clear it by multiplying top and bottom by $i$ , which is allowed because $\frac{i}{i}=1$ . Then use $i^{2}=-1$ , and dividing by $-1$ changes the sign.
$\begin{aligned}(-i)^{-19} &= \frac{1}{i}\cdot\frac{i}{i} \\ &= \frac{i}{i^{2}} \\ &= \frac{i}{-1} \\ &= -i\end{aligned}$

Answer

-i

Compact solution

First rewrite the negative power as a reciprocal.

(-i)^{-19}=\frac{1}{(-i)^{19}}

In the denominator, $-i=(-1)\cdot i$ . Separate the two factors.

(-i)^{19}=(-1)^{19}i^{19}

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{19} &= i^{18}\cdot i \\ &= (i^2)^{9}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{9}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (-1)\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}} \\ &= -i\end{aligned}

Since $19$ is odd, $(-1)^{19}=-1$ . Substitute both reduced factors.

\begin{aligned}(-i)^{19} &= (-1)(-i) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}} \\ &= i\end{aligned}

Now substitute into the reciprocal and simplify.

\begin{aligned}(-i)^{-19} &= \frac{1}{i} \\ &= \frac{1}{i}\cdot\frac{i}{i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{i}{i}=1$}} \\ &= \frac{i}{i^2} \\ &= \frac{i}{(-1)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -i\end{aligned}

Common mistake

Watch outThere are two different minus signs here, one inside the bracket and one in the power. Do the power first and the bracket second, and you will not lose either.

(iv)

27i^{-26}

Solution

Compact solution

The $27$ is a real number multiplying the rest, so leave it outside and work only on $i^{-26}$ . A negative power means turn it upside down.
$27i^{-26} = 27\cdot\frac{1}{i^{26}}$
Work on the bottom. The exponent $26$ is even, so the whole power can be written using $i^{2}$ . Halve it, because $26=2\times 13$ .
$i^{26} = \left(i^{2}\right)^{13}$
Put $-1$ in place of $i^{2}$ .
$i^{26} = (-1)^{13}$
$13$ copies of $-1$ cancel in pairs, and $13$ is odd, so one minus is left over.
$i^{26} = -1$
Now put the bottom back into the whole expression.
$27i^{-26} = 27\cdot\frac{1}{-1}$
$1$ divided by $-1$ is $-1$ , and plus times minus gives minus.
$27i^{-26} = 27\cdot(-1) = -27$

Answer

-27

Compact solution

First rewrite the negative power as a reciprocal.

i^{-26}=\frac{1}{i^{26}}

The exponent is even, so write the power in terms of $i^2$ .

\begin{aligned}i^{26} &= (i^2)^{13} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{13} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}}\end{aligned}

Now substitute into the reciprocal and simplify.

\begin{aligned}i^{-26} &= \frac{1}{-1} \\ &= -1\end{aligned}

Multiply the reduced power by the coefficient outside it.

\begin{aligned}27i^{-26} &= 27(-1) \\ &= -27\end{aligned}

Common mistake

Watch outThe answer is a real number with no

i

in it. An expression made from

i

does not have to stay imaginary.

(v)

i^{11}+i^{5}

Solution

Compact solution

There are two powers joined by a plus sign. Simplify each one on its own, then add. Start with the first. The exponent $11$ is odd, so take one factor of $i$ out, leaving the even power $10$ .
$i^{11} = i^{10}\cdot i$
Halve the even exponent, because $10=2\times 5$ .
$i^{11} = \left(i^{2}\right)^{5}\cdot i$
Put $-1$ in place of $i^{2}$ . Five is odd, so one minus is left over.
$i^{11} = (-1)^{5}\cdot i = -i$
Now the second power. The exponent $5$ is odd, so take one factor of $i$ out and halve the rest.
$i^{5} = \left(i^{2}\right)^{2}\cdot i = (-1)^{2}\cdot i = i$
Put both answers back into the sum.
$i^{11}+i^{5} = (-i)+(i)$
These two terms are opposites. Adding any number to its own negative gives $0$ , just like $-5+5=0$ .
$i^{11}+i^{5} = 0$

Answer

0

Compact solution

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{11} &= i^{10}\cdot i \\ &= (i^2)^{5}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{5}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (-1)\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}} \\ &= -i\end{aligned}

For the other term, take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{5} &= i^{4}\cdot i \\ &= (i^2)^{2}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{2}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

Add the two reduced powers.

\begin{aligned}i^{11}+i^{5} &= (-i)+(i) \\ &= 0 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a+(-a)=0$}}\end{aligned}

(vi)

\left(i^{4}+i^{3}+i^{2}+i\right)^{2}

Solution

Compact solution

Here, simplifying inside the parentheses avoids an unnecessary expansion. Write down each of the four powers, using only $i^{2}=-1$ . The third is $i^{2}\cdot i=-i$ , and the fourth is $\left(i^{2}\right)^{2}=(-1)^{2}=1$ .
$i^{4}=1, \quad i^{3}=-i, \quad i^{2}=-1, \quad i^{1}=i$
Put those four values into the bracket. Keep each one in its own bracket so no minus sign is lost.
$i^{4}+i^{3}+i^{2}+i = 1+(-i)+(-1)+i$
Move them around so the two real numbers are together and the two $i$ terms are together. You may add in any order.
$i^{4}+i^{3}+i^{2}+i = \bigl(1+(-1)\bigr)+\bigl((-i)+i\bigr)$
Take the first group. Adding $-1$ to $1$ gives $0$ , because a number plus its own negative is always $0$ .
$1+(-1) = 0$
Take the second group. Again a number plus its own negative gives $0$ .
$(-i)+i = 0$
So both groups are $0$ , and the whole bracket is $0$ .
$i^{4}+i^{3}+i^{2}+i = 0+0 = 0$
Now use the power that was waiting outside.
$\left(i^{4}+i^{3}+i^{2}+i\right)^{2} = 0^{2}$
And $0\times 0=0$ .
$\left(i^{4}+i^{3}+i^{2}+i\right)^{2} = 0$

Answer

0

Compact solution

First find the third power using $i^2=-1$ .

i^3=i^2\cdot i=(-1)\cdot i=-i

For the fourth power, square $i^2$ .

\begin{aligned}i^4 &= (i^2)^2 \\ &= (-1)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}}\end{aligned}

Substitute these values and group each term with its additive inverse.

\begin{aligned}i^4+i^3+i^2+i &= 1+(-i)+(-1)+i \\ &= \bigl(1+(-1)\bigr)+\bigl((-i)+i\bigr) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{group additive inverses}$}} \\ &= 0+0 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a+(-a)=0$}} \\ &= 0\end{aligned}

Finally, square the value of the bracket.

\begin{aligned}\left(i^4+i^3+i^2+i\right)^2 &= 0^2 \\ &= 0\end{aligned}

Common mistake

Watch outYou did not need to open a square here, so you did not need

(a+b)^{2}=a^{2}+2ab+b^{2}

. Simplifying inside the bracket first turned a complicated-looking question into

0^{2}

(vii)

\left(\frac{i^{8}}{i^{5}}\right)^{-5}

Solution

Compact solution

Sort out the inside of the bracket first. Two powers of the same letter are being divided, so the powers subtract. This is the rule $\frac{a^{m}}{a^{n}}=a^{m-n}$ .
$\frac{i^{8}}{i^{5}} = i^{8-5} = i^{3}$
The exponent $3$ is odd, so take one factor of $i$ out and use $i^{2}=-1$ on the rest.
$i^{3} = i^{2}\cdot i = (-1)\cdot i = -i$
Put that value in place of the bracket.
$\left(\frac{i^{8}}{i^{5}}\right)^{-5} = (-i)^{-5}$
A negative power means turn it upside down, so the bracket goes under $1$ and the power becomes positive.
$(-i)^{-5} = \frac{1}{(-i)^{5}}$
Work on the bottom. Since $-i=(-1)\cdot i$ , raise each factor to the fifth power on its own, using $(ab)^{n}=a^{n}b^{n}$ .
$(-i)^{5} = (-1)^{5}\cdot i^{5}$
$5$ copies of $-1$ cancel in pairs, and $5$ is odd, so one minus is left over.
$(-1)^{5} = -1$
For the other factor, take one $i$ out and halve the rest.
$i^{5} = \left(i^{2}\right)^{2}\cdot i = (-1)^{2}\cdot i = i$
Multiply the two factors, then put the result back at the bottom.
$(-i)^{-5} = \frac{1}{(-1)\cdot i} = \frac{1}{-i}$
Clear the $i$ from the bottom by multiplying top and bottom by $i$ . Look at the third line carefully: $i^{2}=-1$ , so $-i^{2}$ becomes $-(-1)$ , and a minus in front of a minus gives a plus.
$\begin{aligned}\left(\frac{i^{8}}{i^{5}}\right)^{-5} &= \frac{1}{-i}\cdot\frac{i}{i} \\ &= \frac{i}{-i^{2}} \\ &= \frac{i}{-(-1)} \\ &= i\end{aligned}$

Answer

i

Compact solution

Simplify the quotient inside the bracket.

\begin{aligned}\frac{i^{8}}{i^{5}} &= i^{8-5} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{a^m}{a^n}=a^{m-n}\quad(a\neq0)$}} \\ &= i^{3} \\ &= i^2\cdot i \\ &= (-1)\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -i\end{aligned}

Apply the outside negative exponent to this result.

\begin{aligned}\left(\frac{i^{8}}{i^{5}}\right)^{-5} &= (-i)^{-5} \\ &= \frac{1}{(-i)^{5}} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{-n}=\frac{1}{a^n}\quad(a\neq0)$}}\end{aligned}

Separate the sign from the power of $i$ .

(-i)^{5}=(-1)^{5}i^{5}

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{5} &= i^{4}\cdot i \\ &= (i^2)^{2}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{2}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

Since $5$ is odd, $(-1)^{5}=-1$ . Substitute both reduced factors.

\begin{aligned}(-i)^{5} &= (-1)(i) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}} \\ &= -i\end{aligned}

Substitute the denominator and finish the reciprocal.

\begin{aligned}\left(\frac{i^{8}}{i^{5}}\right)^{-5} &= \frac{1}{-i} \\ &= \frac{1}{-i}\cdot\frac{i}{i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{i}{i}=1$}} \\ &= \frac{i}{-i^2} \\ &= \frac{i}{-(-1)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= i\end{aligned}

(viii)

i^{13}\times i^{29}

Solution

Compact solution

Both powers have the same letter $i$ , so join them into one power first. When powers of the same letter are multiplied, the powers add. This is the rule $a^{m}\cdot a^{n}=a^{m+n}$ .
$i^{13}\cdot i^{29} = i^{13+29}$
Do the addition. To add $29$ to $13$ , add $30$ first to get $43$ , then give back the extra $1$ to get $42$ .
$i^{13}\cdot i^{29} = i^{42}$
The exponent $42$ is even, so the whole power can be written using $i^{2}$ . Halve it, because $42=2\times 21$ .
$i^{42} = \left(i^{2}\right)^{21}$
Put $-1$ in place of $i^{2}$ .
$i^{42} = (-1)^{21}$
$21$ copies of $-1$ cancel in pairs, and $21$ is odd, so one minus is left over.
$i^{13}\cdot i^{29} = -1$

Answer

-1

Compact solution

The bases are the same, so add the exponents.

\begin{aligned}i^{13}\cdot i^{29} &= i^{13+29} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^m a^n=a^{m+n}$}} \\ &= i^{42}\end{aligned}

The exponent is even, so write the power in terms of $i^2$ .

\begin{aligned}i^{42} &= (i^2)^{21} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{21} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= -1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}}\end{aligned}

Thus the original product is

i^{13}\cdot i^{29}=-1

Alternative method

Alternative methodWhen you are confident, use the remainder rule. Divide

42

4

: it goes

10

times with remainder

2

, so

i^{42}=i^{2}=-1

Question 2

Write in terms of $i$ .

(i)

2+\sqrt{-4}

Solution

Compact solution

Only the square root needs rewriting. For $a>0$ , the principal square root satisfies $\sqrt{-a}=i\sqrt a$ . Here $a=4$ .
$\sqrt{-4}=i\sqrt4$
The nonnegative square root of $4$ is $2$ , since $2^{2}=4$ .
$\sqrt4=2$
Substitute this value into the imaginary term.
$\sqrt{-4}=i(2)=2i$
Add back the unchanged real term.
$2+\sqrt{-4}=2+2i$

Answer

2+2i

Compact solution

Use the principal-root convention $\sqrt{-a}=i\sqrt a$ for $a>0$ .

\begin{aligned}\sqrt{-4} &= i\sqrt{4} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\sqrt{-a}=i\sqrt a\quad(a>0)$}} \\ &= i\cdot 2 \\ &= 2i\end{aligned}

Put the simplified root back into the original expression.

\begin{aligned}2+\sqrt{-4} &= 2+2i\end{aligned}

Common mistake

Watch outThe two

2

s in the answer are different things. The first is the real part and the second is the imaginary part, so you cannot add them to make

4

(ii)

3-\sqrt{-7}

Solution

Compact solution

Look at the square root, and keep the minus sign in front of it safe for later. Pull the minus sign out of the number as a factor of $-1$ .
$-7 = 7\times(-1)$
Use the principal-root identity $\sqrt{-a}=i\sqrt{a}$ , valid for positive real $a$ . Here $a=7$ .
$\sqrt{-7} = \sqrt{7}\times\sqrt{-1}$
Put $i$ in place of $\sqrt{-1}$ and write it in front. The $\sqrt{7}$ must stay as it is, because $7$ is not a perfect square.
$\sqrt{-7} = i\sqrt{7}$
Put it back, keeping the minus sign that was there from the start.
$3-\sqrt{-7} = 3-i\sqrt{7}$

Answer

3-i\sqrt{7}

Compact solution

Use the principal-root convention $\sqrt{-a}=i\sqrt a$ for $a>0$ .

\begin{aligned}\sqrt{-7} &= i\sqrt{7} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\sqrt{-a}=i\sqrt a\quad(a>0)$}}\end{aligned}

Retain the minus sign in front of the root.

\begin{aligned}3-\sqrt{-7} &= 3-i\sqrt7\end{aligned}

Common mistake

Watch outBoth

i\sqrt{7}

and

\sqrt{7}\,i

mean the same number. Keep

i

visibly outside the radical;

\sqrt{7i}

means something different.

(iii)

\frac{2}{5}+\frac{\sqrt{-16}}{5}

Solution

Compact solution

Only the square root on top of the second fraction needs work. Pull the minus sign out of the number as a factor of $-1$ .
$-16 = 16\times(-1)$
Use $\sqrt{-a}=i\sqrt{a}$ for positive real $a$ . Here the positive factor is $16$ .
$\sqrt{-16} = \sqrt{16}\times\sqrt{-1}$
Take the first factor, because $4\times 4=16$ .
$\sqrt{16} = 4$
Put $i$ in place of $\sqrt{-1}$ and multiply.
$\sqrt{-16} = 4\times i = 4i$
Put the simplified root back.
$\frac{2}{5}+\frac{\sqrt{-16}}{5} = \frac{2}{5}+\frac{4i}{5}$
Both fractions already have the same bottom number $5$ , so add the tops and keep that bottom.
$\frac{2}{5}+\frac{\sqrt{-16}}{5} = \frac{2+4i}{5}$

Answer

\dfrac{2+4i}{5}

Compact solution

Use the principal-root convention $\sqrt{-a}=i\sqrt a$ for $a>0$ .

\begin{aligned}\sqrt{-16} &= i\sqrt{16} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\sqrt{-a}=i\sqrt a\quad(a>0)$}} \\ &= i\cdot 4 \\ &= 4i\end{aligned}

Both fractions have denominator $5$ , so combine their numerators.

\begin{aligned}\frac25+\frac{\sqrt{-16}}5 &= \frac25+\frac{4i}5 \\ &= \frac{2+4i}5 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac ab+\frac cb=\frac{a+c}b$}}\end{aligned}

Other forms

Also correct

\frac{2}{5}+\frac{4}{5}i

is also correct. It shows the real part

\frac{2}{5}

and the imaginary part

\frac{4}{5}

more clearly.

(iv)

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

Solution

Compact solution

The first root has a positive number inside, so leave it alone. Only the second root has a negative number. Pull the minus sign out as a factor of $-1$ .
$-3 = 3\times(-1)$
Use $\sqrt{-a}=i\sqrt{a}$ for positive real $a$ . Here $a=3$ .
$\sqrt{-3} = \sqrt{3}\times\sqrt{-1} = i\sqrt{3}$
Put it back, keeping the minus sign.
$\sqrt{2}-\sqrt{-3} = \sqrt{2}-i\sqrt{3}$

Answer

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

Compact solution

Use the principal-root convention $\sqrt{-a}=i\sqrt a$ for $a>0$ .

\begin{aligned}\sqrt{-3} &= i\sqrt{3} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\sqrt{-a}=i\sqrt a\quad(a>0)$}}\end{aligned}

The real root $\sqrt2$ stays unchanged. Substitute only for the negative root.

\begin{aligned}\sqrt2-\sqrt{-3} &= \sqrt2-i\sqrt3\end{aligned}

Common mistake

Watch outThe answer

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

is a sum of a real term and an imaginary term, not a single like term. Neither radical has a square factor greater than

1

Question 3

Find the values of $x$ and $y$ . Assume $x$ and $y$ are real.

(i)

(2x+5)+(y-3)i=1+2i

Solution

Compact solution

Both sides are already written as a real number plus a real number times $i$ , so we can compare at once. Two complex numbers are equal only when their real parts match and their imaginary parts match. Start with the real parts, which are the pieces with no $i$ on them.
$2x+5=1$
Now the imaginary parts, which are the numbers multiplying $i$ , without the $i$ itself.
$y-3=2$
Take the first equation. To leave $2x$ alone, take $5$ from both sides. This keeps the equation balanced.
$2x+5-5 = 1-5$
On the left $5-5=0$ , so $2x$ is left. On the right $1-5=-4$ , because taking $5$ away from $1$ goes $4$ below $0$ .
$2x = -4$
To leave $x$ alone, divide both sides by $2$ .
$\frac{2x}{2} = \frac{-4}{2}$
On the left the two $2$ s cancel. On the right $-4\div 2=-2$ , because a minus divided by a plus gives a minus.
$x = -2$
Now the second equation. Add $3$ to both sides.
$y-3+3 = 2+3$
On the left $-3+3=0$ , so $y$ is left. On the right $2+3=5$ .
$y = 5$

Answer

x=-2

and

y=5

Compact solution

Since $x,y\in\mathbb R$ , compare the real parts and solve for $x$ .

\begin{aligned}2x+5&=1 \\ 2x+5-5&=1-5 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{subtract }5\text{ on both sides}$}} \\ 2x&=-4 \\ \frac{2x}{2}&=\frac{-4}{2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by }2$}} \\ x&=-2\end{aligned}

Now compare the imaginary coefficients and solve for $y$ .

\begin{aligned}y-3&=2 \\ y-3+3&=2+3 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add }3\text{ on both sides}$}} \\ y&=5\end{aligned}

Check

Check the answer

\begin{aligned}(2(-2)+5)+((5)-3)i &= (-4+5)+(2)i \\ &= 1+2i \quad \checkmark\end{aligned}

Put

x=-2

and

y=5

into the left side. Note

2\times(-2)=-4

, and then

-4+5=1

. It gives the right side exactly.

(ii)

(3x+2)-(4-y)i=5+3i

Solution

Compact solution

Look at the minus sign in front of the second bracket. It belongs to the imaginary part, so do not lose it. First compare the real parts.
$3x+2=5$
Now the imaginary parts. The whole of $-(4-y)$ is what multiplies $i$ , so the minus sign goes with it.
$-(4-y)=3$
Take the first equation. Take $2$ from both sides.
$3x+2-2 = 5-2$
On the left $2-2=0$ . On the right $5-2=3$ .
$3x = 3$
Divide both sides by $3$ .
$\frac{3x}{3} = \frac{3}{3}$
And $3\div 3=1$ .
$x = 1$
Now the second equation. Deal with the minus sign in front of the bracket first. A minus sign in front of a bracket means the whole bracket is multiplied by $-1$ .
$-1\times(4-y) = 3$
Divide both sides by $-1$ to remove it. $3$ divided by $-1$ is $-3$ .
$4-y = -3$
Take $4$ from both sides.
$4-y-4 = -3-4$
On the left $4-4=0$ , so $-y$ is left. On the right, start at $-3$ and go down $4$ more, which reaches $-7$ .
$-y = -7$
Multiply both sides by $-1$ to turn $-y$ into $y$ . This also turns $-7$ into $7$ .
$y = 7$

Answer

x=1

and

y=7

Compact solution

Compare the real parts and solve for $x$ .

\begin{aligned}3x+2&=5 \\ 3x+2-2&=5-2 \\ 3x&=3 \\ \frac{3x}{3}&=\frac33 \\ x&=1\end{aligned}

The imaginary coefficient is the whole expression $-(4-y)$ , including the outside minus sign.

\begin{aligned}-(4-y)&=3 \\ 4-y&=-3 \\ 4-y-4&=-3-4 \\ -y&=-7 \\ (-1)(-y)&=(-1)(-7) \\ y&=7\end{aligned}

Check · Common mistake

Check the answer

\begin{aligned}(3(1)+2)-(4-(7))i &= (3+2)-(-3)i \\ &= 5+3i \quad \checkmark\end{aligned}

Put

x=1

and

y=7

back in. Note

4-7=-3

, and taking away

-3

is the same as adding

3

, which gives

+3i

Watch outA common mistake is to write

4-y=3

, forgetting the minus sign outside the bracket. That gives the wrong answer

y=1

. Deal with that minus sign first.

(iii)

(2+i)x+(1-2i)y=3+4i

Solution

Compact solution

The left side is not yet split into a real part and an imaginary part, so open the brackets first. In the first bracket, the $x$ outside multiplies each term inside: $x$ times $2$ gives $2x$ , and $x$ times $i$ gives $ix$ .
$(2+i)x = 2x+ix$
Open the second bracket the same way: $y$ times $1$ gives $y$ , and $y$ times $-2i$ gives $-2iy$ .
$(1-2i)y = y-2iy$
Put both back into the equation.
$2x+ix+y-2iy = 3+4i$
Collect the two terms with no $i$ into one bracket, and the two terms with $i$ into another.
$(2x+y)+(ix-2iy) = 3+4i$
Take $i$ out of the second bracket as a common factor. Now the left side is a real number plus a real number times $i$ , so we can compare.
$(2x+y)+(x-2y)i = 3+4i$
Two complex numbers are equal only when both parts match. Compare the real parts, and give the result a number so we can use it later.
$2x+y=3 \qquad \cdots(1)$
Now compare the imaginary parts. These two equations are independent, so solving them together determines both real unknowns.
$x-2y=4 \qquad \cdots(2)$
Rearrange equation $(1)$ to get $y$ on its own, by taking $2x$ from both sides. We choose $y$ because it has no number in front, which keeps the working simple.
$y = 3-2x$
Put this in place of $y$ in equation $(2)$ , so only $x$ is left.
$x-2(3-2x)=4$
Open the bracket one term at a time. In the second product, minus times minus gives plus.
$-2\times 3 = -6 \quad\text{and}\quad -2\times(-2x) = +4x$
Write the opened bracket back into the equation.
$x-6+4x = 4$
Collect the two $x$ terms. One $x$ plus four $x$ makes five $x$ .
$x+4x = 5x$
So there is now only one $x$ term.
$5x-6 = 4$
Add $6$ to both sides.
$5x-6+6 = 4+6$
On the left $-6+6=0$ . On the right $4+6=10$ .
$5x = 10$
Divide both sides by $5$ .
$\frac{5x}{5} = \frac{10}{5}$
And $10\div 5=2$ .
$x = 2$
Put $x=2$ back into $y=3-2x$ .
$y = 3-2(2)$
Do the multiplying first, because multiplying comes before taking away.
$2\times 2 = 4$
And taking $4$ away from $3$ goes $1$ below $0$ .
$y = 3-4 = -1$

Answer

x=2

and

y=-1

Compact solution

The left side is not yet split into real and imaginary parts, so open the brackets and regroup.

\begin{aligned}(2+i)x+(1-2i)y &= 3+4i \\ 2x+ix+y-2iy &= 3+4i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{opening the brackets}$}} \\ (2x+y)+(ix-2iy) &= 3+4i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{gathering like terms}$}} \\ (2x+y)+(x-2y)i &= 3+4i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{taking $i$ out as a common factor}$}}\end{aligned}

Now compare the two parts. This gives two equations in two unknowns.

\begin{aligned}2x+y &= 3\qquad\cdots(1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{real parts}$}} \\ x-2y &= 4\qquad\cdots(2) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{imaginary parts}$}}\end{aligned}

Equation $(1)$ carries no number in front of $y$ , so make $y$ the subject.

\begin{aligned}y &= 3-2x\end{aligned}

Put that into equation $(2)$ , so only $x$ is left.

\begin{aligned}x-2(3-2x) &= 4 \\ x-6+4x &= 4 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-2(3-2x)=-6+4x$}} \\ 5x-6 &= 4 \\ 5x &= 10 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add $6$ to both sides}$}} \\ \frac{5x}{5} &= \frac{10}{5} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by $5$}$}} \\ x &= 2\end{aligned}

Put that value back into $y=3-2x$ .

\begin{aligned}y &= 3-2(2) \\ &= 3-4 \\ &= -1\end{aligned}

Check

Check the answer

\begin{aligned}(2+i)(2)+(1-2i)(-1) &= (4+2i)+(-1+2i) \\ &= (4-1)+(2i+2i) \\ &= 3+4i \quad \checkmark\end{aligned}

Put both answers into the left side. Note that

(1-2i)

times

-1

changes both signs, giving

-1+2i

. It gives the right side exactly.

(iv)

(1-i)x+(2+i)y=4-i

Solution

Compact solution

Open the brackets first. In the first bracket, $x$ times $1$ gives $x$ , and $x$ times $-i$ gives $-ix$ .
$(1-i)x = x-ix$
Open the second bracket: $y$ times $2$ gives $2y$ , and $y$ times $i$ gives $iy$ .
$(2+i)y = 2y+iy$
Put both back into the equation.
$x-ix+2y+iy = 4-i$
Collect the terms with no $i$ , then take $i$ out of the other two as a common factor.
$(x+2y)+(-x+y)i = 4-i$
Two complex numbers are equal only when both parts match. Compare the real parts first.
$x+2y=4 \qquad \cdots(1)$
Now compare the imaginary parts. On the right, $-i$ means $-1$ times $i$ , so the imaginary part there is $-1$ .
$-x+y=-1 \qquad \cdots(2)$
Add equation $(1)$ to equation $(2)$ , left side to left side and right side to right side. We choose to add because the $x$ terms are opposites and will disappear.
$(x+2y)+(-x+y) = 4+(-1)$
On the left, $x$ and $-x$ are opposites, so they cancel.
$x+(-x) = 0$
The other terms on the left join up: two $y$ plus one $y$ makes three $y$ .
$2y+y = 3y$
On the right, adding $-1$ to $4$ is the same as taking $1$ from $4$ .
$4+(-1) = 3$
So the added equation becomes one simple statement about $y$ .
$3y = 3$
Divide both sides by $3$ .
$\frac{3y}{3} = \frac{3}{3}$
And $3\div 3=1$ .
$y = 1$
Put $y=1$ back into equation $(1)$ .
$x+2(1) = 4$
And $2\times 1=2$ .
$x+2 = 4$
Take $2$ from both sides.
$x+2-2 = 4-2$
On the left $2-2=0$ . On the right $4-2=2$ .
$x = 2$

Answer

x=2

and

y=1

Compact solution

Open the brackets and regroup so both sides are in the form $(\text{real})+(\text{real})i$ .

\begin{aligned}(1-i)x+(2+i)y &= 4-i \\ x-ix+2y+iy &= 4-i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{opening the brackets}$}} \\ (x+2y)+(-ix+iy) &= 4-i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{gathering like terms}$}} \\ (x+2y)+(-x+y)i &= 4-i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{taking $i$ out as a common factor}$}}\end{aligned}

Comparing the two parts gives two equations.

\begin{aligned}x+2y &= 4\qquad\cdots(1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{real parts}$}} \\ -x+y &= -1\qquad\cdots(2) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{imaginary parts}$}}\end{aligned}

The $x$ terms are opposites, so adding the two equations removes $x$ at once.

\begin{aligned}(x+2y)+(-x+y) &= 4+(-1) \\ 3y &= 3 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{the $x$ terms cancel}$}} \\ \frac{3y}{3} &= \frac33 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by $3$}$}} \\ y &= 1\end{aligned}

Put that value back into equation $(1)$ .

\begin{aligned}x+2(1) &= 4 \\ x+2 &= 4 \\ x &= 2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{take $2$ from both sides}$}}\end{aligned}

Check

Check the answer

\begin{aligned}(1-i)(2)+(2+i)(1) &= (2-2i)+(2+i) \\ &= (2+2)+(-2i+i) \\ &= 4-i \quad \checkmark\end{aligned}

Put both answers in. In the last line

-2i+i=-i

, because taking away two

i

and adding one back leaves one

i

still owing.

(v)

(3x-1)+(2y-3)i=8+7i

Solution

Compact solution

Both sides are already split into a real part and an imaginary part, so compare them directly. Two complex numbers are equal only when both parts match. Start with the real parts.
$3x-1=8$
Now the imaginary parts, which are the numbers multiplying $i$ .
$2y-3=7$
Take the first equation. Add $1$ to both sides.
$3x-1+1 = 8+1$
On the left $-1+1=0$ . On the right $8+1=9$ .
$3x = 9$
Divide both sides by $3$ .
$\frac{3x}{3} = \frac{9}{3}$
And $9\div 3=3$ .
$x = 3$
Now the second equation. Add $3$ to both sides.
$2y-3+3 = 7+3$
On the left $-3+3=0$ . On the right $7+3=10$ .
$2y = 10$
Divide both sides by $2$ .
$\frac{2y}{2} = \frac{10}{2}$
And $10\div 2=5$ .
$y = 5$

Answer

x=3

and

y=5

Compact solution

Since $x,y\in\mathbb R$ , compare the real parts and solve for $x$ .

\begin{aligned}3x-1&=8 \\ 3x-1+1&=8+1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add }1\text{ on both sides}$}} \\ 3x&=9 \\ \frac{3x}{3}&=\frac{9}{3} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by }3$}} \\ x&=3\end{aligned}

Now compare the imaginary coefficients and solve for $y$ .

\begin{aligned}2y-3&=7 \\ 2y-3+3&=7+3 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add }3\text{ on both sides}$}} \\ 2y&=10 \\ \frac{2y}{2}&=\frac{10}{2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by }2$}} \\ y&=5\end{aligned}

Check

Check the answer

\begin{aligned}(3(3)-1)+(2(5)-3)i &= (9-1)+(10-3)i \\ &= 8+7i \quad \checkmark\end{aligned}

Put

x=3

and

y=5

into the left side. It gives the right side exactly.

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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 imaginary unit

Complex numbers and rectangular form

Real and purely imaginary complex numbers

Equality of complex numbers

Why we needed a new kind of number

Powers of ii

Negative powers and the reciprocal of ii

Square roots of negative numbers

From the number line to the complex plane

Example 1

Example 2

Question 1

Question 2

Question 3

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

Powers of $i$

Negative powers and the reciprocal of $i$

Class 10
Mathematics