# Subtraction 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 mini-lesson, we will learn how to subtract complex numbers and calculate complex numbers. Check-out 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 Do You Mean by Subtraction of 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 \(z\).

Subtraction of complex numbers means doing the mathematical operation of subtraction on complex numbers.

**Complex Plane**

Every complex number indicates a point in the XY-plane.

For example, the complex number \(x+iy\) represents the point \((x,y)\) in the XY-plane.

Look at the plane shown below.

The points \((2,4), (-2,3), (0,1), (2,0), (-5,-2)\), and \((1,-2)\) corresponds to the complex numbers \(2+4i, -2+3i,0+i1,2+0i,-5-2i\), and \(1-2i\) respectively.

The plane where a complex number is assigned to each of its points is called a complex plane.

**Parallelogram Law of Subtraction of Complex Numbers**

The subtraction of complex numbers is geometrically similar to vector subtraction.

Let's see how to subtract complex numbers by using parallelogram law.

Take two complex numbers \(z_{1}=a_{1}+ib_{1}\) and \(z_{1}=a_{1}+ib_{1}\).

Then plot the two complex numbers in the complex plane, and consider them as normal vectors.

Now, we need to calculate \(z_{1}-z_{2}\).

For that we need to add the two vectors, \(z_{1}\) and \(-z_{2}\).

To find \(-z_{2}\), vector reverse the \(z_{2}\) vector with same magnitude.

Now, add the two vectors, \(z_{1}\) and \(-z_{2}\) by using the parallelogram law of vector addition which is shown below.

The resultant vector represents the vector \(z_{1}-z_{2}\) and its tip represent the point \(z_{1}-z_{2}\).

Let's visualize the subtraction of complex numbers by using parallelogram law with the help of the simulation given below.

Drag the tip of the two complex vectors to visualize the parallelogram law of subtraction for different complex numbers.

**What Are the Steps to Subtract Complex Numbers?**

We have already learned how to subtract complex numbers geometrically.

But isn't that process a little confusing and tough?

Let's discuss an easy process to subtract the two complex numbers.

To subtract the two complex numbers follow the steps:

- Group the real numbers of the two complex numbers and imaginary numbers of the complex numbers.
- Then simplify them by applying corresponding mathematical operations.

**Example**

Let us subtract complex number \(z_{2}=3+4i\) from the complex number \(z_{1}=5+7i\) by using these steps.

\[\begin{aligned}z_{2}&=3+4i\\z_{1}&=5+7i\\z_{1}-z_{2}&=(5+7i)-(3+4i)\end{aligned}\]

Now simplify by using mathematical operations:

\[\begin{aligned}z_{1}-z_{2}&=(5-3)+(7i-4i)\\&=2+3i\end{aligned}\]

Thus, the subtraction of the given two complex numbers is: \[z_{1}-z_{2}= 2+3i\]

- All imaginary numbers are complex numbers but all complex numbers don't need to be imaginary numbers.
- All real numbers are complex numbers but all complex numbers don't need to be real numbers.
- The conjugate of the complex \(z=a+ib\) is \(\overline{z}=a-ib\).
- The difference of two complex numbers satisfies the closure law because if \(z_{1}\) and \(z_{2}\) are two complex numbers, then \(z_{1}-z_{2}\) is also a complex number.
- The difference of two complex numbers do not satisfies the commutative law because if \(z_{1}\) and \(z_{2}\) are two complex numbers, then \(z_{1}-z_{2} \neq z_{2}-z_{1}\).

**Solved Examples**

Example 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 imaginary part of the difference of complex number \(-2+3i\) and its conjugate."

Can you guess the lucky number?

**Solution**

The conjugate of the complex number \(a+ib\) is \(a-ib\).

So, the conjugate of the complex number \(-2+3i\) is \(-2-3i\).

Let's subtract complex numbers, \(-2+3i\) and \(-2-3i\).

\[\begin{align}(-2+3i)-(-2-3i)&=-2+3i+2+3i\\&=6i\end{align}\]

The required number is 6 |

Example 2 |

Find the roots of the quadratic equation \(\sqrt{5}x^2+x+\sqrt{5}=0\).

What is the difference between both the roots?

**Solution**

The discriminant of the quadratic equation \(ax^2+bx+c=0\) is given by \(D=b^2-4ac\).

So, let's find the discriminant.

\[\begin{align}D&=b^2-4ac\\&=1^2-4(\sqrt{5})^2\\&=1-20\\&=-19\end{align}\]

The roots of the quadratic equation \(ax^2+bx+c=0\) is given by \(x=\dfrac{-b \pm \sqrt{D}}{2a}\).

\[\begin{align}x&=\dfrac{-b \pm \sqrt{D}}{2a}\\&=\dfrac{-1 \pm \sqrt{-19}}{2\sqrt{5}}\\&=\dfrac{-1 \pm \sqrt{19}i}{2\sqrt{5}}\end{align}\]

The difference between \(\dfrac{-1+\sqrt{19}}{2\sqrt{5}}\) and \(\dfrac{-1 - \sqrt{19}}{2\sqrt{5}}\) is \(\sqrt{\dfrac{19}{5}}\)

So, the difference of the roots is \(\sqrt{\dfrac{19}{5}}\). |

- What is the square of the difference of two complex numbers, \(z_{1}\) and \(z_{2}\)?
- What is the difference between a complex number and its conjugate?

**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 mini-lesson 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.

**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 learning-teaching-learning 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 Angle Sum Theorem**

### 1. Is 0 a complex number?

Yes, 0 is a complex number part with the real part and imaginary part as 0

### 2. How do you write a complex number?

The complex number is written in the form of \(a+ib\), where \(a\) and \(b\) are real numbers.

### 3. Is the set of complex numbers closed under subtraction?

Yes, the set of complex numbers closed under subtraction because if \(z_{1}\) and \(z_{2}\) are two complex numbers, then \(z_{1}-z_{2}\) is also a complex number.

### 4. How do you simplify complex numbers?

We simplify complex numbers by using the fact that \(i^2=1\) and considering them as binomials.

For example, the product of \(3+i\) and \(1+2i\) is shown below.

\[ \begin{align}(3+i)(1+2i)&= 3+6i+i+2i^2\\[0.2cm] &= 3+7i-2 \\[0.2cm] &=1+7i \end{align} \]

### 5. How do you expand complex numbers?

We expand complex numbers by the fact that \(i^2=1\) and considering them as binomials.

For example, the expansion of the product of \((3+i)(1+2i)\) is shown below.

\[ \begin{align} (3+i)(1+2i)&= 3+6i+i+2i^2\\[0.2cm] &= 3+7i-2 \\[0.2cm] &=1+7i \end{align} \]

### 6. How do you add two complex numbers?

The two complex numbers \(z_{1}=a_{1}+ib_{1}\) and \(z_{2}=a_{2}+ib_{2}\) can be added as \(z_{1}+z_{2}=(a_{1}+a_{2})+i(b_{1}+b_{2})\).

### 7. How do you find the difference between complex numbers?

The difference of two complex numbers \(z_{1}=a_{1}+ib_{1}\) and \(z_{2}=a_{2}+ib_{2}\) is given by \(z_{1}-z_{2}=(a_{1}-a_{2})+i(b_{1}-b_{2})\).

### 8. How do you subtract complex numbers from polar form?

Convert the complex numbers into polar form to subtract complex numbers from the polar form.

### 9. What is the product of two complex numbers?

We multiply complex numbers by considering them as binomials.

\[ \begin{align}(3+i)(1+2i)&= 3+6i+i+2i^2\\[0.2cm] &= 3+7i-2 \\[0.2cm] &=1+7i \end{align} \]

### 10. How do you square a complex number?

The square of a complex number \(z=a+ib\) is given by \(z^{2}=(a+ib)^2=(a^2-b^2)+i(2ab)\).