Division of complex Numbers
We already know the quadratic formula to solve a quadratic equation.
While doing this, sometimes, the value inside the square root may be negative.
For example, while solving a quadratic equation x^{2} + x + 1 = 0 using the quadratic formula, we get:
So far we know that the square roots of negative numbers are NOT real numbers.
Then what type of numbers are they?
They are complex numbers.
In this minilesson, we will learn about the division of complex numbers, division of complex numbers in polar form, the division of imaginary numbers, and dividing complex fractions. Checkout the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.
Lesson Plan
What Are Complex Numbers?
We denote \(\sqrt{1}\) by the symbol \(i\) which we call "iota".
So, we have i^{2} = 1
This means you can say that \(i\) is the solution of the quadratic equation x^{2} + 1 = 0
The number of the form z=a+ib, where \(a\) and \(b\) are real numbers are called the complex numbers.
Here, \(a\) is called the real part, and \(b\) is called the imaginary part of the complex number \(z\).
Division of complex numbers means doing the mathematical operation of division on complex numbers.
Complex Numbers in Polar Form
Every complex number can also be written in polar form.
The graphical representation of the complex number \(a+ib\) is shown in the graph below.
The parameters \(r\) and \(\theta\) are the parameters of the polar form.
The polar form of the complex number \(z=a+ib\) is given by: \(z=r\left(\cos\theta+i\sin\theta\right)\)
where \(r\) is the modulus (\(z\)) of the complex number and \(\theta\) is the argument of the complex number.
What Do You Mean by Division of Complex Numbers?
The division of complex numbers is mathematically similar to the division of two real numbers.
If \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\) are the two complex numbers.
Then the division of two complex numbers is mathematically written as:
\[\dfrac{z_1}{z_2}=\dfrac{x_1+iy_1}{x_2+iy_2}\]
Division of Complex Numbers Formula
The division of two complex numbers \(z_1=a+ib\) and \(z_2=c+id\) is given by the quotient \(\dfrac{a+ib}{c+id}\).
This is calculated by using the division of complex numbers formula:
\[\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{ac+bd}{c^2+d^2}+i\left(\dfrac{bcad}{c^2+d^2}\right)\end{aligned}\] 
What Are the Steps to Divide Complex Numbers?
We have already learned how to divide complex numbers.
Now let's discuss the steps on how to divide the complex numbers.
To divide the two complex numbers follow the steps:
 First, calculate the conjugate of the complex number that is at the denominator of the fraction.
 Multiply the conjugate with the numerator and the denominator of the complex fraction.
 Apply the algebraic identity \((a+b)(ab)=a^2b^2\) in the denominator and substitute \(i^2=1\).
 Apply the distributive property in the numerator and simplify.
 Separate the real part and the imaginary part of the resultant complex number.
\[\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{a+ib}{c+id}\\&=\dfrac{a+ib}{c+id}\times\dfrac{cid}{cid}\\&=\dfrac{(a+ib)(cid)}{c^2+(id)^2}\\&=\dfrac{aciad+ibci^2bd}{c^2(1)d^2}\\&=\dfrac{aciad+ibc+bd}{c^2+d^2}\\&=\dfrac{(ac+bd)+i(bcad)}{c^2+d^2}\\&=\dfrac{ac+bd}{c^2+d^2}+i\left(\dfrac{bcad}{c^2+d^2}\right)\end{aligned}\]
Experiment with the simulation given below to divide two complex numbers by changing the sliders for \(a, b, c\) and \(d\).
Division of Complex Numbers in Polar Form
Let us divide the complex number \(z_{1}=r_1\left(\cos\theta_1+i\sin\theta_1\right)\) by the complex number \(z_{2}=r_2\left(\cos\theta_2+i\sin\theta_2\right)\).
The division of complex numbers in polar form is calculated as:
\[\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{r_1\left(\cos\theta_1+i\sin\theta_1\right)}{r_2\left(\cos\theta_2+i\sin\theta_2\right)}\\&=\dfrac{r_1\left(\cos\theta_1+i\sin\theta_1\right)}{r_2\left(\cos\theta_2+i\sin\theta_2\right)}\left(\dfrac{\cos\theta_2i\sin\theta_2}{\cos\theta_2i\sin\theta_2}\right)\\&=\dfrac{r_1\left(\cos\theta_1+i\sin\theta_1\right)\left(\cos\theta_2i\sin\theta_2\right)}{r_2\left(\cos^2\theta_2(i)^2\sin^2\theta_2\right)}\\&=\dfrac{r_1\left(\cos\theta_1+i\sin\theta_1\right)\left(\cos\theta_2i\sin\theta_2\right)}{r_2(\cos^2\theta_2+\sin^2\theta_2)}\\&=\frac{r_1}{r_2}\left[\cos(\theta_1\theta_2)+i\sin(\theta_1\theta_2)\right]\\&=r\left(\cos\theta+i\sin\theta\right)\end{aligned}\]
Where \(\theta=\theta_1\theta_2\) and \(r=\dfrac{r_1}{r_2}\).
Thus, the division of complex numbers \(z_{1}=r_1\left(\cos\theta_1+i\sin\theta_1\right)\) and \(z_{2}=r_2\left(\cos\theta_2+i\sin\theta_2\right)\) in polar form is given by the quotient \(\dfrac{r_1\left(\cos\theta_1+i\sin\theta_1\right)}{r_2\left(\cos\theta_2+i\sin\theta_2\right)}\).
This is calculated by the formula:
\[\begin{aligned}\dfrac{z_1}{z_2}&=r\left(\cos\theta+i\sin\theta\right)\end{aligned}\] 
Here, \(\theta=\theta_1\theta_2\) and \(r=\dfrac{r_1}{r_2}\).
 To divide a complex number \(a+ib\) by \(c+id\), multiply the numerator and denominator of the fraction \(\dfrac{a+ib}{c+id}\) by \(cid\) and simplify.
 The conjugate of the complex \(z=a+ib\) is \(\overline{z}=aib\).
 The modulus of the complex number \(z=a+ib\) is \(z=\sqrt{a^2+b^2}\).
Solved Examples
Example 1 
Convert the complex number \(z=1+i\sqrt{3}\) in the polar form.
Represent this complex number on a complex plane.
Solution
Let \(a=1\) and \(b=\sqrt{3}\).
The complex number \(z=1+i\sqrt{3}\) is plotted in the graph shown below.
The value of \(r\) is given by:
\[\begin{align}r&=\sqrt{a^2+b^2}\\&=\sqrt{1+3}\\&=4 \end{align}\]
Now, let's find the argument \(\theta\).
\[\begin{align}\theta&=\tan^{1}\left(\dfrac{b}{a}\right)\\&=\tan^{1}\left(\dfrac{\sqrt{3}}{1}\right)\\&=\dfrac{\pi}{3}\end{align}\]
\(\therefore\) The polar form is \(z=2\left(\cos\left(\dfrac{\pi}{3}\right)+i\sin\left(\dfrac{\pi}{3}\right)\right)\). 
Example 2 
Jolly asked Emma to express the complex number \(\dfrac{5+\sqrt{2}i}{1\sqrt{2}i}\) in the form of \(a+ib\).
Can you help them?
Solution
Let \(a=5\), \(b=\sqrt{2}\), \(c=1\), and \(d=\sqrt{2}\).
\[\begin{aligned}\dfrac{a+ib}{c+id}&=\dfrac{ac+bd}{c^2+d^2}+i\left(\dfrac{bcad}{c^2+d^2}\right)\\&=\dfrac{(5\times 1)+(\sqrt{2}\times(\sqrt{2}))}{1^2+(\sqrt{2})^2}+\\&i\left(\dfrac{(\sqrt{2} \times 1)(5\times(\sqrt{2}))}{1^2+(\sqrt{2})^2}\right)\\&=\dfrac{52}{1+2}+i\left(\dfrac{\sqrt{2}+5\sqrt{2}}{1+2}\right)\\&=\dfrac{3+6\sqrt{2}i}{1+2}\\&=1+2\sqrt{2}i\end{aligned}\]
\(\dfrac{5+\sqrt{2}i}{1\sqrt{2}i}=1+2\sqrt{2}i\) 
Example 3 
Jake is stuck with one question in his maths assignment.
The question is to find the resultant complex number by dividing \(3+4i\) by \(82i\).
Can you help him?
Solution
We will find simlify the complex number \(\dfrac{3+4i}{82i}\).
The conjugate of the denominator \(82i\) is \(8+2i\).
Multiply the numerator and denominator of \(\dfrac{3+4i}{82i}\) by \(8+2i\).
\[\begin{align}\dfrac{3+4i}{82i}&=\dfrac{3+4i}{82i}\times\dfrac{8+2i}{8+2i}\\&=\dfrac{24+6i+32i+8i^2}{64+16i16i4i^2}\\&=\dfrac{16+38i}{68}\\&=\dfrac{4}{17}+\dfrac{19}{34}i\end{align}\]
\(\dfrac{3+4i}{82i}=\dfrac{4}{17}+\dfrac{19}{34}i\) 
1. 
Frank has a secret lucky number with him. He gives a few hints to his friend Joe to identify it. 


He says "It is the resultant complex number by dividing \(3+4i\) by \(43i\)." Is the lucky number a real number or an imaginary number? 
Interactive Questions
Here are a few activities for you to practice.
Select/Type your answer and click the "Check Answer" button to see the result.
Let's Summarize
The minilesson targeted the fascinating concept of the subtraction of complex numbers. The math journey around the subtraction of complex numbers starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
We hope you understood the division of complex numbers, division of complex numbers in polar form, the division of imaginary numbers, and dividing complex fractions along with the concept of division of complex numbers.
About Cuemath
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learningteachinglearning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
FAQs on Division of Complex Numbers
1. How to divide complex numbers?
The division of two complex numbers \(z_1=a+ib\) and \(z_2=c+id\) is calculated by using the division of complex numbers formula:
\[\dfrac{z_1}{z_2}=\dfrac{ac+bd}{c^2+d^2}+i\left(\dfrac{bcad}{c^2+d^2}\right)\]
2. How do you write a complex number?
A complex number is written in the form of \(a+ib\), where \(a\) and \(b\) are real numbers.
3. What is the symbol of a complex number?
The symbol of the complex number is \(z\).
4. How do you solve complex numbers?
Keep in mind the following points while solving the complex numbers:
 While adding and subtracting the complex numbers, group the real part and the imaginary parts together.
 While multiplying the two complex numbers, use the value \(i^2=1\)
 While dividing the complex numbers, multiply the fraction with the conjugate of the denominator.
5. Is 6 a complex number?
Yes, the number 6 is a complex number whose imaginary part is zero.
6. How do you write a quotient as a complex number?
Let the quotient be \(\dfrac{a+ib}{c+id}\).
This can be written as \(\dfrac{ac+bd}{c^2+d^2}+i\left(\dfrac{bcad}{c^2+d^2}\right)\).
7. What is the quotient when 4+8i is divided by 1+3i?
The quotient \(\dfrac{4+8i}{1+3i}\) is given as \(\dfrac{14}{5}i\dfrac{2}{5}\).
8. How do you divide a complex number by a real number?
Divide the real part and the imaginary part of the complex number by that real number separately.
9. How do you divide square roots with complex numbers?
To divide the square root with complex number use the substitution \(i=\sqrt{1}\).
For example:
\[ \begin{align}\frac{\sqrt{2}}{i}&=\frac{\sqrt{2}}{\sqrt{1}}\\[0.2cm] &=\sqrt{\frac{2}{1}}\\[0.2cm] &=\sqrt{2}\end{align} \]
10. What is the modulus of a complex number?
The modulus of a complex number \(z=a+ib\) is given by \(z=a^2+b^2\).