## Divide Complex Numbers

### Learning Outcomes

• Identify and write the complex conjugate of a complex number.
• Divide complex numbers.
• Simplify powers of $i$.

## Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of $a+bi$ is $a-bi$.

Note that complex conjugates have a reciprocal relationship: The complex conjugate of $a+bi$ is $a-bi$, and the complex conjugate of $a-bi$ is $a+bi$. Importantly, complex conjugate pairs have a special property. Their product is always real.

\begin{align}(a+bi)(a-bi)&=a^2-abi+abi-b^2i^2\\[2mm]&=a^2-b^2(-1)\\[2mm]&=a^2+b^2\end{align}

Suppose we want to divide $c+di$ by $a+bi$, where neither $a$ nor $b$ equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

$\dfrac{c+di}{a+bi}$ where $a\ne 0$ and $b\ne 0$.

Multiply the numerator and denominator by the complex conjugate of the denominator.

$\dfrac{\left(c+di\right)}{\left(a+bi\right)}\cdot \dfrac{\left(a-bi\right)}{\left(a-bi\right)}=\dfrac{\left(c+di\right)\left(a-bi\right)}{\left(a+bi\right)\left(a-bi\right)}$

Apply the distributive property.

$=\dfrac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}$

Simplify, remembering that ${i}^{2}=-1$.

\begin{align}&=\dfrac{ca-cbi+adi-bd\left(-1\right)}{{a}^{2}-abi+abi-{b}^{2}\left(-1\right)} \\[2mm] &=\dfrac{\left(ca+bd\right)+\left(ad-cb\right)i}{{a}^{2}+{b}^{2}}\end{align}

### A General Note: The Complex Conjugate

The complex conjugate of a complex number $a+bi$ is $a-bi$. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.

### Example: Finding Complex Conjugates

Find the complex conjugate of each number.

1. $2+i\sqrt{5}$
2. $-\frac{1}{2}i$

### How To: Given two complex numbers, divide one by the other.

1. Write the division problem as a fraction.
2. Determine the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.

### Example: Dividing Complex Numbers

Divide $\left(2+5i\right)$ by $\left(4-i\right)$.

### Example: Substituting a Complex Number into a Polynomial Function

Let $f\left(x\right)={x}^{2}-5x+2$. Evaluate $f\left(3+i\right)$.

### Try It

Let $f\left(x\right)=2{x}^{2}-3x$. Evaluate $f\left(8-i\right)$.

### Example: Substituting an Imaginary Number in a Rational Function

Let $f\left(x\right)=\dfrac{2+x}{x+3}$. Evaluate $f\left(10i\right)$.

### Try It

Let $f\left(x\right)=\dfrac{x+1}{x - 4}$. Evaluate $f\left(-i\right)$.

## Simplifying Powers of $i$

The powers of $i$ are cyclic. Let’s look at what happens when we raise $i$ to increasing powers.

${i}^{1}=i$
${i}^{2}=-1$
${i}^{3}={i}^{2}\cdot i=-1\cdot i=-i$
${i}^{4}={i}^{3}\cdot i=-i\cdot i=-{i}^{2}=-\left(-1\right)=1$
${i}^{5}={i}^{4}\cdot i=1\cdot i=i$

We can see that when we get to the fifth power of $i$, it is equal to the first power. As we continue to multiply $i$ by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of $i$.

${i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1$
${i}^{7}={i}^{6}\cdot i={i}^{2}\cdot i={i}^{3}=-i$
${i}^{8}={i}^{7}\cdot i={i}^{3}\cdot i={i}^{4}=1$
${i}^{9}={i}^{8}\cdot i={i}^{4}\cdot i={i}^{5}=i$

### Example: Simplifying Powers of $i$

Evaluate ${i}^{35}$.

### Q & A

Can we write ${i}^{35}$ in other helpful ways?

As we saw in Example: Simplifying Powers of $i$, we reduced ${i}^{35}$ to ${i}^{3}$ by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of ${i}^{35}$ may be more useful. The table below shows some other possible factorizations.

 Factorization of ${i}^{35}$ ${i}^{34}\cdot i$ ${i}^{33}\cdot {i}^{2}$ ${i}^{31}\cdot {i}^{4}$ ${i}^{19}\cdot {i}^{16}$ Reduced form ${\left({i}^{2}\right)}^{17}\cdot i$ ${i}^{33}\cdot \left(-1\right)$ ${i}^{31}\cdot 1$ ${i}^{19}\cdot {\left({i}^{4}\right)}^{4}$ Simplified form ${\left(-1\right)}^{17}\cdot i$ $-{i}^{33}$ ${i}^{31}$ ${i}^{19}$

Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.

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