# Algebra 2 - Multiplying and Dividing Complex Numbers

## Multiplying Complex Numbers

• Multiplication of complex numbers is accomplished in a manner similar to the multiplication of binomials. We can use either the distributive property or the FOIL [multiplying First, Outer, Inner, and Last terms together] method to perform the multiplication operation.
• For the sake of simplicity, we can split the multiplication of complex numbers into three different cases.

1) Multiplying a complex number by a real number

• $k·\left(a+ib\right)=ka+ikb$, where $k$ is any real number.

2) Multiplying a complex number by a pure imaginary number

• $\left(ci\right)·\left(a+ib\right)=aci+{i}^{2}bc=aci-bc=-bc+iac$, where $ci$ is a pure imaginary number.

3) Multiplying two complex numbers

• $\left(a+ib\right)·\left(c+id\right)=$$a\left(c+id\right)+ib\left(c+id\right)=$$ac+iad+ibc+{i}^{2}bd=$$ac+iad+ibc-bd=$$\left(ac-bd\right)+i\left(ad+bc\right)$

## Dividing Complex Numbers

• It is difficult to divide a complex number by another complex number because we cannot divide anything by an imaginary number. The division is only possible when the fraction has a real-number denominator.
• So, the first step towards division must be to eliminate the complex part in the denominator. To do so, we multiply the numerator and the denominator by the complex conjugate of the denominator.
• Recall that the complex conjugate of $z=a+ib$ is $\overline{)z}=a-ib$.
• Suppose we want to divide ${z}_{1}=a+bi$ by ${z}_{2=}c+di$, where ${z}_{2}\ne 0$. We first write the division as a fraction, find the complex conjugate of the denominator, and multiply it by the numerator and denominator of the fraction.

$\frac{{z}_{1}}{{z}_{2}}=\frac{\left(a+ib\right)}{\left(c+id\right)}=$$\frac{\left(a+ib\right)}{\left(c+id\right)}×\frac{\left(c-id\right)}{\left(c-id\right)}=$$\frac{a·\left(c-id\right)+ib·\left(c-id\right)}{\left(c+id\right)·\left(c-id\right)}$

$=\frac{ac-iad+ibc-{i}^{2}bd}{{c}^{2}+{i}^{2}{d}^{2}}=$$\frac{ac-iad+ibc+bd}{{c}^{2}-{d}^{2}}=$$\frac{\left(ac+bd\right)+i\left(bc-ad\right)}{{c}^{2}-{d}^{2}}=$$\frac{\left(ac+bd\right)}{\left({c}^{2}-{d}^{2}\right)}+i\frac{\left(bc-ad\right)}{\left({c}^{2}-{d}^{2}\right)}$

## Solved Examples

Example 1: Find the product of $\left(1-4i\right)$ and $\left(5-2i\right)$.

Solution: $\left(1-4i\right)·\left(5-2i\right)=$$\left(1\right)·\left(5-2i\right)-\left(4i\right)·\left(5-2i\right)=$$5-2i-20i+8{i}^{2}=$$5-22i-8=$$-3-22i$

Example 2: Compute $\left(7+i\right)·\left(1-5i\right)$.

Solution: $\left(7+i\right)·\left(1-5i\right)=$$\left(7\right)·\left(1-5i\right)+\left(i\right)·\left(1-5i\right)=$$7-35i+i-5{i}^{2}=$$7-34i+5=$$12-34i$

Example 3: Express ${\left(2+i\right)}^{2}$ in the standard form of a complex number.

Solution: ${\left(2+i\right)}^{2}=\left(2+i\right)·\left(2+i\right)=$$\left(2\right)·\left(2+i\right)+\left(i\right)·\left(2+i\right)=$$4+2i+2i+{i}^{2}=$$4+4i-1=$$3+4i$

Example 4: Find the quotient of $\left(8-3i\right)$ and $\left(5+2i\right)$.

Solution: $\frac{\left(8-3i\right)}{\left(5+2i\right)}=$$\frac{\left(8-3i\right)}{\left(5+2i\right)}×\frac{\left(5-2i\right)}{\left(5-2i\right)}=$$\frac{\left(8-3i\right)·\left(5-2i\right)}{{5}^{2}-{\left(2i\right)}^{2}}$

$=\frac{\left(8\right)·\left(5-2i\right)-\left(3i\right)·\left(5-2i\right)}{{5}^{2}-4{i}^{2}}=$$\frac{40-16i-15i+6{i}^{2}}{25+4}$

$=\frac{40-31i-6}{29}=\frac{34-31i}{29}=\frac{34}{29}-\frac{31}{29}i$

Example 5: Simplify $\frac{3}{7i}$.

Solution: The denominator can be written as $\left(0+7i\right)$. So, the complex conjugate of the denominator will be $\left(0-7i\right)$ or simply $-7i$.

$\frac{3}{7i}×\frac{\left(-7i\right)}{\left(-7i\right)}=$$\frac{-21i}{-49{i}^{2}}=$$\frac{-21i}{49}=-\frac{21}{49}i$

## Cheat Sheet

• Complex number multiplication is carried out similarly to the multiplication of binomials.
• In complex number division, we first multiply the numerator and denominator by the complex conjugate of the denominator and then simplify it until we reach the standard form $a+ib$.
• The complex conjugate of $z=a+ib$ is $\overline{)z}=a-ib$.

## Blunder Areas

• It should be noted that all real numbers are complex numbers.
• The complex conjugate of any pure imaginary number is negative of the same number—for example, the complex conjugate of $3i$ is $-3i$.