Complex numbers: Perform operations on complex numbers

# Unit 1: Basic complex number arithmetic

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Perform addition, subtraction and multiplication on complex numbers in standard/rectangular form.
• Perform division on complex numbers in standard form introducing the concept of conjugate.

## What you should know

Before you start this unit, make sure you can:

## Introduction

In subject outcome 1.1, we were introduced to imaginary numbers and complex numbers in standard or rectangular form. We know that complex numbers in standard form are represented as $\scriptsize a+bi$. But to be really useful, we need to be able to work with complex numbers in the same ways as we work with real numbers. We need to simplify expressions, solve equations and plot functions.

One example of this is when we work with capacitors and inductors in alternating current (AC) electric circuits. Components like inductors and capacitors provide a kind of resistance to the flow of AC called inductive and capacitive reactance respectively. The amount of reactance depends on the voltage ($\scriptsize \text{V}$) and current ($\scriptsize \text{I}$) and so can be described using complex numbers where $\scriptsize z=\text{V}+i\text{I}$. In order to find the total reactance in a circuit, we need to add up all these reactances.

Thankfully, we can perform the four basic arithmetic operations of addition, subtraction, multiplication and division on complex numbers just like we do with real numbers.

## Adding and subtracting complex numbers

To add or subtract complex numbers, we use a technique that you will have seen before when doing algebra – collecting like terms together. To add or subtract complex numbers, we combine the real parts together and then combine the imaginary parts together.

### Example 1.1

1. $\scriptsize (3+4i)+(2\text{ }+\text{ }5i)$
2. $\scriptsize (-5\text{ }+\text{ }7i)+(-11\text{ }+\text{ }2i)$
3. $\scriptsize (-3-4i)+(4-2i)$

Solutions

1. .
\scriptsize \begin{align*}(3+4i)+(2+5i)&=3+4i+2+5i&&\text{Write the complex numbers without brackets}\\& =(3+2)+(4+5)I&&\text{Collect the real and imaginary parts together}\\&=5+9i\end{align*}
2. .
\scriptsize \begin{align*} (-5+7i)+(-11+2i) & =-5+7i-11+2i\\ & =(-5-11)+(7+2)i\\ & =-16+9i\end{align*}
3. .
\scriptsize \begin{align*} (-3-4i)+(4-2i) & =-3-4i+4-2i\\ & =(-3+4)+(-4-2)i\\ & =1-6i\end{align*}

### Example 1.2

Subtract the following complex numbers:

1. $\scriptsize (3-4i)-(2+5i)$
2. $\scriptsize (8+3i)-(4-7i)$
3. $\scriptsize (7-4i)-(-3-9i)$

Solutions

1. .
\scriptsize \begin{align*} (3-4i)-(2+5i)&=3-4i-2-5i &&\text{Write the complex numbers without brackets.}\\ &&&\text{Be careful of the negative sign.}\\ &=(3-2)+(-4-5)i &&\text{Collect the real and imaginary parts.}\\ &=1-9i \end{align*}
2. .
\scriptsize \begin{align*} (8+3i)-(4-7i)& =8+3i-4+7i && \text{Be careful of the signs when writing the complex}\\&&&\text {numbers without brackets.}\\ & =(8-4)+(3+7)i\\ & =4+10i\end{align*}
3. .
\scriptsize \begin{align*}(7-4i)-(-3-9i)&=7-4i+3+9i\\&=(7+3)+(-4+9)I\\&=10+5i\end{align*}

$\scriptsize (a+bi)+(c+di)=(a+c)+(b+d)i$

Subtracting complex numbers:
$\scriptsize (a+bi)-(c+di)=(a-c)+(b-d)i$

### Note

If you have an internet connection, watch this excellent summary of adding and subtracting complex numbers called “Ex 1: Adding and Subtracting Complex Numbers”.

### Exercise 1.1

1. $\scriptsize (-3+2i)-(-2+5i)$
2. $\scriptsize (-2+4i)+(-2-4i)$
3. $\scriptsize (8+6i)-(-6-6i)$
4. $\scriptsize (5+3i)-(2-5i)+(3+4i)-(4+2i)$

The full solutions are at the end of the unit.

## Multiplying complex numbers

The process of multiplying complex numbers is very similar to multiplying two binomials. The only differences are that we always work with the real and imaginary parts separately, and we need to remember that $\scriptsize {{i}^{2}}=-1$.

### Example 1.3

Simplify the following:

1. $\scriptsize 3(6+2i)$
2. $\scriptsize 2i(5-3i)$
3. $\scriptsize (4+3i)(2-5i)$
4. $\scriptsize {{(2-i)}^{3}}$

Solutions

1. When multiplying a complex number by a real number, we distribute the real number as we would with a binomial.
\scriptsize \begin{align*}3(6+2i)\\=18+6i\end{align*}
2. When multiplying a complex number by an imaginary number, we distribute the imaginary number as we would with a binomial.
\scriptsize \begin{align*}&2i(5-3i)\\&=10i-6({{i}^{2}})\\&=10i-6(-1)&& \text{Remember that }{{i}^{2}}=-1\\&=10i+6\\&=6+10i && \text{Standard form is written as }a+bi\end{align*}
3. When we multiply two complex numbers together, we use the same process as when multiplying two binomials. Remember that FOIL is an acronym for First, Outer, Inner, and Last.
\scriptsize \begin{align*}(4+3i)(2-5i)&=8-20i+6i-15({{i}^{2}})\\&=8-20i+6i-15(-1)&& \text{Remember that }{{i}^{2}}=-1\\&=8-20i+6i+15 &&\text{Group the real and imaginary parts}\\&=23-14i\end{align*}
4. $\scriptsize {{(2-i)}^{3}}=(2-i)(2-i)(2-i)$
We start by multiplying the first two complex numbers together.
\scriptsize \begin{align*}(2-i)(2-i)(2-i)&=(4-2i-2i+{{i}^{2}})(2-i)&& \text{Do the multiplication in brackets}\\&=(4-4i-1)(2-i)&&({{i}^{2}}=-1)\\&=(3-4i)(2-i)\end{align*}
Now we can multiply the product of the first two complex numbers by the third complex number.
\scriptsize \begin{align*}(3-4i)(2-i)&=6-3i-8i+4{{i}^{2}}\\&=6-3i-8i-4&& \text{Group the real and imaginary parts}\\&=2-11i\end{align*}

Multiplying complex numbers:

\scriptsize \begin{align*}(a+bi)(c+di)&=ac+adi+bci+bd{{i}^{2}}\\&=ac+adi+bci-bd&& ({{i}^{2}}=-1)\\&=(ac-bd)+(ad+bc)i\end{align*}

### Note

If you have an internet connection, watch the video “Ex 2: Multiply Complex Numbers” for more on multiplying complex numbers.

### Example 1.4

Simplify $\scriptsize (5-4{{i}^{5}})(6+3{{i}^{7}})$.

Solution

\scriptsize \begin{align*}(5-4{{i}^{5}})(6+3{{i}^{7}})&=30+15{{i}^{7}}-24{{i}^{5}}-12{{i}^{{12}}}\\&=30+15.{{i}^{4}}.{{i}^{3}}-24.{{i}^{4}}.i-12.{{i}^{{4\times 3}}}&& \text{Factor out as many factors of }{{i}^{4}}\text{ as possible}\\&=30+15.1.{{i}^{3}}-24.1.i-{{12.1}^{3}}&& \text{Remember that }{{i}^{4}}=1\\&=30+15.(-1).i-24i-12\\&=30-15i-24i-12\\&=18-39i\end{align*}

### Exercise 1.2

1. $\scriptsize (-3+7i)(-3-7i)$
2. $\scriptsize (2\text{ }-\sqrt{{-16}})(2+\sqrt{{-9}})$
3. $\scriptsize (12\sqrt{{-2}})(-4\sqrt{{-4}}+1)$
4. $\scriptsize (-1+i)(3+4i)(3i)$
5. $\scriptsize (1-i)(1+i)(2-3i)$
6. $\scriptsize i(2+3{{i}^{7}})+(-2-i)$

The full solutions are at the end of the unit.

## Dividing complex numbers

Unfortunately, dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number. This means that any fraction must have a real-number denominator so that we can write the answer in standard form $\scriptsize a+bi$.

We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with only a real number as the denominator. This is called the complex conjugate of the denominator. Thankfully finding the complex conjugate is very easy. We just change the sign. In other words, the complex conjugate of $\scriptsize a+bi$ is $\scriptsize a-bi$.

Have a look at what happens when we multiply $\scriptsize a+bi$ by its complex conjugate $\scriptsize a-bi$:

\scriptsize \begin{align*}&(a+bi)(a-bi)\\&={{a}^{2}}-abi+abi-{{b}^{2}}{{i}^{2}}\\&={{a}^{2}}-{{b}^{2}}{{i}^{2}}\\&={{a}^{2}}+{{b}^{2}}\end{align*}

We have eliminated the imaginary part of the complex number.

### Take note!

We can only find the complex conjugate of a complex number if it is written in standard form. If the complex number is not in standard form, first write it in standard form.

### Exercise 1.3

Find the complex conjugates of the following:

1. $\scriptsize 2+3i$
2. $\scriptsize -3-4i$
3. $\scriptsize 3+\sqrt{5}i$
4. $\scriptsize -\displaystyle \frac{1}{2}i$

The full solutions are at the end of the unit.

### Take note!

The complex conjugate of the complex number $\scriptsize a+bi$ is $\scriptsize a-bi$. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.

### Example 1.5

Divide $\scriptsize (2+5i)$ by $\scriptsize 4-i$.

Solution

Step 1: If necessary, write the problem as a fraction, making sure that both the numerator and denominator are in standard form.
$\scriptsize \displaystyle \frac{{(2+5i)}}{{(4-i)}}$

Step 2: Multiply the numerator and denominator by the complex conjugate of the denominator.
$\scriptsize \displaystyle \frac{{(2+5i)}}{{(4-i)}}\times \displaystyle \frac{{(4+i)}}{{(4+i)}}$

Step 3: Multiply and simplify
\scriptsize \begin{align*}\displaystyle \frac{{(2+5i)}}{{(4-i)}}\times \displaystyle \frac{{(4+i)}}{{(4+i)}}&=\displaystyle \frac{{8+2i+20i+5{{i}^{2}}}}{{16+4i-4i-{{i}^{2}}}}\\&=\displaystyle \frac{{8+2i+20i-5}}{{16-(-1)}}\\&=\displaystyle \frac{{3+22i}}{{17}}\end{align*}

Step 4: Separate the real and imaginary parts to write the answer in standard form.
$\scriptsize \displaystyle \frac{3}{{17}}+\displaystyle \frac{{22}}{{17}}i$

Dividing complex numbers:

\scriptsize \begin{align*}&\displaystyle \frac{{a+bi}}{{c+di}}\quad \text{where }c\ne 0\text{ and }d\ne 0\\&=\displaystyle \frac{{(a+bi)}}{{(c+di)}}\times \displaystyle \frac{{(c-di)}}{{(c-di)}}\end{align*}

### Exercise 1.4

Simplify the following, writing the answer in standard form:

1. $\scriptsize \displaystyle \frac{{(2+3i)}}{{(4+3i)}}$
2. $\scriptsize \displaystyle \frac{{6+4i}}{i}$
3. $\scriptsize \displaystyle \frac{{7-\sqrt{{-5}}}}{{6+\sqrt{{-4}}}}$
4. $\scriptsize \displaystyle \frac{{5-2i}}{{(3+4i)(2-i)}}$
5. $\scriptsize \displaystyle \frac{{{{{\left( {2i} \right)}}^{5}}}}{{4i}}$
6. $\scriptsize \displaystyle \frac{{(1+3i)(2-4i)}}{{(1+2i)}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to add, subtract and multiply complex numbers.
• The complex conjugate of $\scriptsize a+bi$ is $\scriptsize a-bi$.
• How to divide complex numbers by multiplying the numerator and denominator by the complex conjugate of the denominator.

# Unit 1: Assessment

#### Suggested time to complete: 40 minutes

1. $\scriptsize -(10+4i)-(7+4i)$
2. $\scriptsize (-3+{{i}^{3}})(2-3{{i}^{3}})$
3. $\scriptsize (-i)(4+2i+{{i}^{2}})+3i+2$
4. $\scriptsize \displaystyle \frac{{24{{i}^{8}}-36{{i}^{{16}}}+48{{i}^{{25}}}}}{{12{{i}^{8}}}}$
5. $\scriptsize \displaystyle \frac{{{{{\left( {3+i} \right)}}^{2}}}}{{{{{\left( {1+2i} \right)}}^{2}}}}$
6. $\scriptsize \displaystyle \frac{{3+2i}}{{1+2i}}-\displaystyle \frac{{2-3i}}{{3+i}}$
2. Evaluate $\scriptsize 2{{x}^{2}}+x-3$ if $\scriptsize x=2-3i$.

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
\scriptsize \begin{align*} (-3+2i)-(-2+5i)&=-3+2i+2-5i\\ &=(-3+2)+(2-5)i\\ &=-1-3i \end{align*}
2. .
\scriptsize \begin{align*}(-2+4i)+(-2-4i)&=-2+4i-2-4i\\&=(-2-2)+(4-4)i\\&=-4+0i\end{align*}
3. .
\scriptsize \begin{align*}(8+6i)-(-6-6i)&=8+6i+6+6i\\&=(8+6)+(6+6)i\\&=14+12i\end{align*}
4. .
\scriptsize \begin{align*} (5+3i)-(2-5i)+(3+4i)-(4+2i)&=5+3i-2+5i+3+4i-4-2i\\ &=(5-2+3-4)+(3+5+4-2)i\\ &=2+10i \end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. .
\scriptsize \begin{align*} (-3+7i)(-3-7i)&=9+21i-21i-49i^2\\ &=9+49\\ &=58+0i \end{align*}
2. .
\scriptsize \begin{align*}(2-\sqrt{{-16}})(2+\sqrt{{-9}})&=(2-4i)(2+3i)\\&=4+6i-8i-12{{i}^{2}}\\&=4-2i+12\\&=16-2i\end{align*}
3. .
\scriptsize \begin{align*}(12\sqrt{{-2}})(-4\sqrt{{-4}}+1)&=(12\sqrt{2}i)(-4(2)i+1)\\&=(12\sqrt{2}i)(-8i+1)\\&=-96\sqrt{2}{{i}^{2}}+12\sqrt{2}i\\&=96\sqrt{2}+12\sqrt{2}i\end{align*}
4. .
\scriptsize \begin{align*} (-1+i)(3+4i)(3i)&=(-3-4i+3i+4i^2)(3i)\\ &=(-3-i-4)(3i)\\ &=(-7-i)(3i)\\ &=-21i-3i^2\\ &=3-21i \end{align*}
5. .
\scriptsize \begin{align*} (1-i)(1+i)(2-3i)&=(1+i-i-i^2)(2-3i)\\ &=(1+1)(2-3i)\\ &=2(2-3i)\\ &=4-6i \end{align*}
6. .
\scriptsize \begin{align*} i(2+3{{i}^{7}})+(-2-i)&=2i+3{{i}^{8}}-2-i\\&=2i+3{{({{i}^{4}})}^{2}}-2-i\\&=2i+3{{(1)}^{2}}-2-i\\&=2i+3-2-i\\&=1+i\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. Complex conjugate of $\scriptsize 2+3i$ is $\scriptsize 2-3i$
2. Complex conjugate of $\scriptsize -3-4i$ is $\scriptsize -3+4i$
3. Complex conjugate of $\scriptsize 3+\sqrt{5}i$ is $\scriptsize 3-\sqrt{5}i$
4. Write the number in standard form first: $\scriptsize 0-\displaystyle \frac{1}{2}i$
.
The complex conjugate of $\scriptsize 0-\displaystyle \frac{1}{2}i$ is $\scriptsize 0+\displaystyle \frac{1}{2}i=\displaystyle \frac{1}{2}i$

Back to Exercise 1.3

### Exercise 1.4

1. .
\scriptsize \begin{align*} \displaystyle \frac{(2+3i)}{(4+3i)}&=\displaystyle \frac{(2+3i)}{(4+3i)}\times \displaystyle \frac{(4-3i)}{(4-3i)}\\ &=\displaystyle \frac{8-6i+12i-9i^2}{16-12i+12i-9i^2}\\ &=\displaystyle \frac{8-6i+12i+9}{16+9}\\ &=\displaystyle \frac{17+6i}{25}\\ &=\displaystyle \frac{17}{25}+\displaystyle \frac{6}{25}i \end{align*}
2. .
\scriptsize \begin{align*} \displaystyle \frac{6+4i}{i}&=\displaystyle \frac{(6+4i)}{(0+i)}\\ &=\displaystyle \frac{(6+4i)}{(0+i)}\times \displaystyle \frac{0-i}{0-i}\\ &=\displaystyle \frac{-6i-4i^2}{-i^2}\\ &=\displaystyle \frac{-6i+4}{1}\\ &=4-6i \end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{{7-\sqrt{{-5}}}}{{6+\sqrt{{-4}}}}&=\displaystyle \frac{{7-\sqrt{5}i}}{{6+2i}}\\&=\displaystyle \frac{{\left( {7-\sqrt{5}i} \right)}}{{\left( {6+2i} \right)}}\times \displaystyle \frac{{\left( {6-2i} \right)}}{{\left( {6-2i} \right)}}\\&=\displaystyle \frac{{42-14i-6\sqrt{5}i+2\sqrt{5}{{i}^{2}}}}{{36-12i+12i-4{{i}^{2}}}}\\&=\displaystyle \frac{{42-14i-6\sqrt{5}i-2\sqrt{5}}}{{36+4}}\\&=\displaystyle \frac{{42-\left( {14+6\sqrt{5}} \right)i-2\sqrt{5}}}{{40}}\\&=\displaystyle \frac{{42-2\sqrt{5}}}{{40}}-\displaystyle \frac{{14+6\sqrt{5}}}{{40}}i\\&=\displaystyle \frac{{21-\sqrt{5}}}{{20}}-\displaystyle \frac{{7+3\sqrt{5}}}{2}i\end{align*}
4. .
\scriptsize \begin{align*} \displaystyle \frac{5-2i}{(3+4i)(2-i)}&=\displaystyle \frac{(5-2i)}{(6-3i+8i-4i^2}\\ &=\displaystyle \frac{(5-2i)}{10+5i}\\ &=\displaystyle \frac{(5-2i)}{10+5i}\times \displaystyle \frac{10-5i}{10-5i}\\ &=\displaystyle \frac{50-25i-20i+10i^2}{100-50i+50i-25i^2}\\ &=\displaystyle \frac{40-45i}{125}\\ &=\displaystyle \frac{40}{125}-\displaystyle \frac{45}{125}i\\ &=\displaystyle \frac{8}{25}-\displaystyle \frac{9}{25}i \end{align*}
5. .
\scriptsize \begin{align*} \displaystyle \frac{(2i)^5}{4i}&=\displaystyle \frac{32i^5}{4i}\\ &=8i^4\\ &=8 \end{align*}
Or
\scriptsize \begin{align*}\displaystyle \frac{{{{{\left( {2i} \right)}}^{5}}}}{{4i}}&=\displaystyle \frac{{32{{i}^{5}}}}{{4i}}\\&=\displaystyle \frac{{32{{i}^{5}}}}{{0+4i}}\\&=\displaystyle \frac{{32{{i}^{5}}}}{{0+4i}}\times \displaystyle \frac{{(0-4i)}}{{(0-4i)}}\\&=-\displaystyle \frac{{-128{{i}^{6}}}}{{16{{i}^{2}}}}\\&=\displaystyle \frac{{-128.{{i}^{4}}.{{i}^{2}}}}{{16}}\\&=\displaystyle \frac{{128}}{{16}}=8\end{align*}
6. .
\scriptsize \begin{align*} \displaystyle \frac{(1+3i)(2-4i)}{(1+2i)}&=\displaystyle \frac{2-4i+6i-12i^2}{(1+2i)}\\ &=\displaystyle \frac{(14+2i)}{(1+2i)}\times \displaystyle \frac{1-2i}{1-2i}\\ &=\displaystyle \frac{14-28i+2i-4i^2}{1-4i^2}\\ &=\displaystyle \frac{18-26i}{5}\\ &=\displaystyle \frac{18}{5}-\displaystyle \frac{26}{5}i \end{align*}

Back to Exercise 1.4

### Unit 1: Assessment

1. .
1. .
\scriptsize \begin{align*} -(10+4i)-(7+4i)&=-10-4i-7-4i\\ &=-17-8i \end{align*}
2. .
\scriptsize \begin{align*}(-3+{{i}^{3}})(2-3{{i}^{3}})&=-6+9{{i}^{3}}+2{{i}^{3}}-3{{i}^{6}}\\&=-6+9{{i}^{3}}+2{{i}^{3}}-3.{{i}^{4}}.{{i}^{2}}\\&=-6+9{{i}^{3}}+2{{i}^{3}}+3\\&=-3+11{{i}^{3}}\\&=-3-11i\end{align*}
3. .
\scriptsize \begin{align*}(-i)(4+2i+{{i}^{2}})+3i+2&=-4i-2{{i}^{2}}-{{i}^{3}}+3i+2\\&=-4i+2+i+3i+2\\&=4+0i\\&=4\end{align*}
4. .
\scriptsize \begin{align*} \displaystyle \frac{24i^8-36i^{16}+48i^{25}}{12i^8}&=\displaystyle \frac{24(i^4)^2-36(i^4)^4+48(i^4)^6.i}{12(i^4)^2}\\ &=\displaystyle \frac{24-36+48i}{12}\\ &=\displaystyle \frac{-12+48i}{12}\\ &=\displaystyle \frac{-12}{12}+\displaystyle \frac{48}{12}i\\ &=-1+4i \end{align*}
5. .
\scriptsize \begin{align*} \displaystyle \frac{(3+i)^2}{(1+2i)^2}&=\displaystyle \frac{(3+i)(3+i)}{(1+2i)(1+2i)}\\ &=\displaystyle \frac{9+6i-1}{1+4i-4}\\ &=\displaystyle \frac{(8+6i)}{(-3+4i)}\times \displaystyle \frac{-3-4i}{-3-4i}\\ &=\displaystyle \frac{-24-32i-18i-24i^2}{9-16i^2}\\ &=\displaystyle \frac{0-50i}{25}\\ &=0-2i \end{align*}
6. .
\scriptsize \begin{align*} \displaystyle \frac{3+2i}{1+2i}-\displaystyle \frac{2-3i}{3+i}&=\displaystyle \frac{(3+2i)}{(1+2i)}\times \displaystyle \frac{(1-2i)}{(1-2i)}-\displaystyle \frac{(2-3i)}{(3+i)}\times \displaystyle \frac{(3-i)}{(3-i)}\\ &=\displaystyle \frac{3-6i+2i-4i^2}{1-4i^2}-\displaystyle \frac{6-2i-9i+3i^2}{9-i^2}\\ &=\displaystyle \frac{7-4i}{5}-\displaystyle \frac{3-11i}{10}\\ &=\displaystyle \frac{2(7-4i)-(3-11i)}{10}\\ &=\displaystyle \frac{14-8i-3+11i}{10}\\ &=\displaystyle \frac{11+3i}{10}\\ &=\displaystyle \frac{11}{10}+\displaystyle \frac{3}{10}i \end{align*}
2. $\scriptsize 2{{x}^{2}}+x-3$ if $\scriptsize x=2-3i$
\scriptsize \begin{align*} 2{{(2-3i)}^{2}}+(2-3i)-3&=2(2-3i)(2-3i)+2-3i-3\\&=2(4-6i-6i+9{{i}^{2}})-1-3i\\&=2(-5-12i)-1-3i\\&=-10-24i-1-3i\\&=-11-27i\end{align*}

Back to Unit 1: Assessment