Operations with Complex Numbers – Examples and Practice Problems

A complex number is defined as the combination of a real number and an imaginary number. Real numbers are the numbers that we use in everyday life and with which we perform mathematical calculations. Imaginary numbers are numbers that contain the imaginary unit, which is defined as the square root of negative one.

The basic operations that we can perform with complex numbers and that we will see here are:

  • Addition of complex numbers
  • Subtraction of complex numbers
  • Multiplication of complex numbers
  • Division of complex numbers
ALGEBRA
operations with complex numbers

Relevant for

Learning to perform operations with complex numbers.

See examples

ALGEBRA
operations with complex numbers

Relevant for

Learning to perform operations with complex numbers.

See examples

Addition of complex numbers

We know that a complex number has the form $latex z=a+bi$, where a and b are real numbers.

Let’s consider two complex numbers $latex z_{1} = a_{1}+b_{1}i$ and $latex z_{2}=a_{2}+b_{2}i$. Therefore, the sum of these complex numbers is defined as:

$latex z_{1}+z_{2}=(a_{1}+a_{2}) + (b_{1}+b_{2})i$

This means that the real part of the resulting complex number is equal to the sum of the real parts of each complex number and the imaginary part of the resulting complex number is equal to the sum of the imaginary parts of each complex number.

If we have sums of more than two complex numbers, the same idea is applied. The real part of the result is equal to the sum of all the real parts of each number and the imaginary part is equal to the sum of all the imaginary parts of each number.

EXAMPLES

  • We have the numbers $latex z_{1}=16-28i$ and $latex z_{2}=9+12i$. Find the result of adding them.

We know that the resulting complex number has the form $latex z=a+bi$, where is the sum of the real parts of the numbers and is the sum of imaginary parts of numbers:

$latex a=16+9=25$

$latex b=-28+12=-16$

⇒   $latex z=25-16i$

  • We have the numbers $latex z_{1}=a+5i$, $latex z_{2}=5+bi$ and $latex z_{3}=10+8i$. Find the value of a and b if we have $latex z_{3}=z_{1}+z_{2}$.

We know that the real part of the resulting number is the sum of the real parts of the numbers $latex z_{1}$ and $latex z_{2}$ and the imaginary part is the sum of the imaginary parts of the numbers $latex z_{1}$ and $latex z_{2}$. Therefore, we have:

$latex 10=a+5$

⇒  $latex a=5$

$latex 8=5+b$

⇒  $latex b=3$

Try solving the following exercise

If we have the numbers $latex z_{1}=a+4i$, $latex z_{2}=-3+bi$ and $latex z_{3}=6+10i$, find the value of a and b if $latex z_{3}=z_{1}+z_{2}$.

Choose an answer







Subtraction of complex numbers

Subtraction of complex numbers is very similar to complex number addition in that we subtract the real parts of the numbers and the imaginary parts separately. Considering the complex numbers $latex z_{1}= a_{1}+b_{1} i$ and $latex z_{2}=a_{2}+b_{2}i$. Therefore, the difference $latex z_{1} -z_{2}$ is defined as:

$latex z_{1}-z_{2}=(a_{1}-a_{2})+(b_{1}-b_{2})i$

EXAMPLES

  • If we have $latex z_{1}=12-7i$ and $latex z_{2}=7+5i$, solve the subtraction $latex z_{1}-z_{2}$.

We know that we have to subtract the real and imaginary parts separately and that the resulting number has the form $latex z=a+bi$. Therefore, we have:

$latex a=12-7=5$

$latex b=-7-5=-12$

⇒   $latex z=5-12i$

  • We have the numbers $latex z_{1}=a+7i$, $latex z_{2}=2+bi$ and $latex z_{3}=5+4i$. Find the values of a and b if we have $latex z_{3}=z_{1}-z_{2}$.

We have to subtract the real and imaginary parts separately, then we have:

$latex 5=a-2$

⇒  $latex a=7$

$latex 4=7-b$

⇒  $latex b=3$

Try solving the following exercise

If we have the numbers $latex z_{1}=5+ai$, $latex z_{2}=2+3i$ and $latex z_{3}=3$, find the value of a if we have $latex z_{3}=z_{1}-z_{2}$.

Choose an answer







Multiplication of complex numbers

We know the expansion $latex (a+b)(c+d)=ac+ad+bc+bd$. In the same way, if we have the complex numbers $latex z_{1}=a_{1}+b_{1}i$ and $latex z_{2}=a_{2}+b_{2}i$, their product is equal to:

$latex z_{1}z_{2}=(a_{1}+b_{1}i)(a_{2}+b_{2}i)$

$latex =a_{1}a_{2}+a_{1}b_{2}i+b_{1}a_{2}i+b_{1}b_{2}{{i}^2}$

We know that $latex {{i}^2}=-1$, therefore, we have:

$latex z_{1}z_{2}=(a_{1}a_{2}-b_{1}b_{2})+(a_{1}b_{2}+a_{2}b_{1})i$

EXAMPLES

  • Multiply the numbers $latex z_{1}=3+5i$ and $latex z_{2}=2+4i$.

We use the process shown:

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

$$=(3)(2)+(5i)(2)+(3)(4i)+(5i)(4i)$$

$latex =6+10i+12i+20{{i}^2}$

$latex =6+22i-20$

$latex =-14+22i$

  • Multiply the numbers $latex z_{1}=4-2i$ and $latex z_{2}=-3+3i$.

We use the process shown:

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

$$=(4)(-3)+(-2i)(-3)+(4)(3i)+(-2i)(3i)$$

$latex =-12+6i+12i-6{{i}^2}$

$latex =-12+18i+6$

$latex =-6+18i$

Try solving the following exercise

Multiply the numbers $latex z_{1}=6-2i$ and $latex z_{2}=4+3i$.

Choose an answer







Division of complex numbers

To divide complex numbers, we have to multiply both the numerator and denominator by the conjugate of the complex number in the denominator. To find the conjugate, we simply change the sign that is between the two terms of the denominator.

Then we distribute the multiplication and simplify. Specifically, we have to remember that $latex {{i}^2}=-1$. Finally, we write the answer in the form $latex a+bi$.

EXAMPLES

  • Solve the division $latex \frac{4+5i}{2-3i}$.

We have to multiply both the denominator and the numerator by the conjugate of the denominator which is equal to $latex 2+3i$:

$$\frac{4+5i}{2-3i}=\frac{4+5i}{2-3i} \times \frac{2+3i}{2+3i}$$

$$=\frac{8+12i+10i+15{{i}^2}}{4-6i+6i-9{{i}^2}}$$

$$=\frac{8+22i-15}{4+9}$$

$$=\frac{-7+22i}{13}$$

We already got the answer, but we have to write it in the form $latex a+bi$:

$$=\frac{-7}{13}+\frac{22}{13}i$$

  • Solve the division $latex \frac{6-3i}{4+2i}$.

We multiply both the denominator and the numerator by the conjugate of the denominator which is equal to $latex 4-2i$:

$$\frac{6-3i}{4+2i}=\frac{6-3i}{4+2i} \times \frac{4-2i}{4-2i}$$

$$=\frac{24-12i-12i+6{{i}^2}}{16+8i-8i-4{{i}^2}}$$

$$=\frac{24-24i-6}{16+4}$$

$$=\frac{18-24i}{20}$$

We already got the answer, but we have to write it in the form $latex a+bi$:

$$ =\frac{18}{20}-\frac{24}{20}i$$

$$=\frac{9}{10}-\frac{6}{5}i$$

Try solving the following exercise

Solve the division $latex \frac{3+2i}{4-3i}$.

Choose an answer







See also

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