Complex numbers are written in the general form *a+bi*, where *a* and *b* are real numbers and “*i*” is the imaginary unit that is equal to the square root of negative one. Basic operations like addition, subtraction, multiplication, and division can be done with complex numbers. The main idea when performing operations with complex numbers is to separate their real and imaginary parts.

Here, we will learn how to solve the subtraction of complex numbers. In addition, we will look at several examples with answers to master this topic completely.

## How to solve subtraction of complex numbers?

To solve subtractions of complex numbers, we have to identify their real and imaginary parts and subtract them separately. This is very similar to subtracting polynomials, where we identify and subtract like terms.

Therefore, assuming we have the complex numbers $latex z_{1}=a+bi$ and $latex z_{2}=c+di$, their subtraction is equal to:

$latex z_{1}-z_{2}=(a-c)+(b-d)i$ |

We can see that the real part of the result is equal to the subtraction of the real parts of the original numbers and the imaginary part of the result is equal to the subtraction of the imaginary parts of the original numbers. This means that we have:

$latex Re(z_{1}-z_{2})=Re(z_{1})-Re(z_{2})$

$latex Im(z_{1}-z_{2})=Im(z_{1})-Re(z_{2})$

We can subtract any number of complex numbers by following this simple rule.

## Subtraction of complex numbers – Examples with answers

With the following examples, you can apply what you have learned about the subtraction of complex numbers. Each example has its respective solution, but it is recommended that you try to solve the exercises yourself before looking at the answer.

**EXAMPLE 1**

If we have the numbers $latex z_{1}=4+5i$ and $latex z_{2}=3+6i$, what is the result of $latex z_{1}-z_{2}$?

##### Solution

We identify the real and imaginary parts of the numbers and subtract them separately. Therefore, we have:

$latex z_{1}-z_{2}=(4-3)+(5-6)i$

$latex =1-i$

**EXAMPLE 2**

If we have the numbers $latex z_{1}=6+8i$ and $latex z_{2}=5-7i$, what is the result of $latex z_{1}-z_{2}$?

##### Solution

We identify the real and imaginary parts of the numbers and subtract them separately. Therefore, we have:

$latex z_{1}-z_{2}=(6-5)+(8-(-7))i$

$latex =1+15i$

**EXAMPLE 3**

Solve the subtraction $latex z_{1}-z_{2}-z_{3}$ if we have the numbers $latex z_{1}=3+6$, $latex z_{2}=4-5$ and $latex z_{3}=-2+7$.

##### Solution

Here, we have three numbers, but the process to follow is the same. We group the real and imaginary parts and subtract them separately:

$$z_{1}-z_{2}-z_{3}=(3-4-(-2))+(6-(-5)-7)i$$

$latex =1+4i$

**EXAMPLE 4**

Solve the subtraction $latex z_{1}-z_{2}-z_{3}$ if we have the numbers $latex z_{1}=10-8$, $latex z_{2}=-6+7$ and $latex z_{3}=3-10$.

##### Solution

Similar to the previous exercise, we simply have to group the real and imaginary parts and subtract them separately:

$$z_{1}-z_{2}-z_{3}=(10-(-6)-3)+(-8-7-(-10))i$$

$latex =13-5i$

**EXAMPLE 5**

If we have the complex numbers $latex z_{1}=a+8i$, $latex z_{2}=-5+bi$ and $latex z_{3}=10+5i$, what is the value of *a* and *b* if we have $latex z_{3}=z_{1}-z_{2}$?

##### Solution

If we subtract the real and imaginary parts of the numbers $latex z_{1}$ and $latex z_{2}$ separately, we have:

$latex 10=a-(-5)$

⇒ $latex a=5$

$latex 5=8-b$

⇒ $latex b=3$

**EXAMPLE 6**

If we have the complex numbers $latex z_{1}=a-6i$, $latex z_{2}=7+bi$ and $latex z_{3}=-5-10i$, what is the value of *a* and *b* if we have $latex z_{3}=z_{1}-z_{2}$?

##### Solution

If we subtract the real and imaginary parts of the numbers $latex z_{1}$ and $latex z_{2}$ separately, we have:

$latex -5=a-7$

⇒ $latex a=2$

$latex -10=-6-b$

⇒ $latex b=4$

## Subtraction of complex numbers – Practice problems

Put into practice what you have learned to solve the following complex number subtraction problems. Solve the subtractions and select your answer. Click “Check” to verify that you got the correct answer.

## See also

Interested in learning more about operations with complex numbers? Take a look at these pages:

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