### Learning Outcomes

- Identify and write the complex conjugate of a complex number.
- Divide complex numbers.
- Simplify powers of [latex]i[/latex].

### recall a technique: rationalizing the denominator

In mathematics, a technique that works well in one situation often works equally as well in a different, but stylistically similar situation.

Dividing complex numbers uses a technique you used when rewriting a fraction that contains a radical in the denominator. We called it *rationalizing a denominator. *To accomplish it, you multiplied the fraction in the numerator and denominator by a number that would clear the radical in the denominator. We use the same general idea when dividing complex numbers. This time, instead of multiplying a radical by a specially chosen radical though, we’ll multiply a complex number by a specially chosen complex number.

When rationalizing the denominator, we took advantage of the fact that [latex]\left(\sqrt{a}\right)^2=a[/latex].

When dividing complex numbers, we’ll take advantage of the fact that [latex]i^2 = -1[/latex].

## 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 [latex]a+bi[/latex] is [latex]a-bi[/latex].

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

[latex]\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}[/latex]

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

[latex]\dfrac{c+di}{a+bi}[/latex] where [latex]a\ne 0[/latex] and [latex]b\ne 0[/latex].

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

[latex]\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)}[/latex]

Apply the distributive property.

[latex]=\dfrac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[/latex]

Simplify, remembering that [latex]{i}^{2}=-1[/latex].

[latex]\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}[/latex]

### A General Note: The Complex Conjugate

The **complex conjugate** of a complex number [latex]a+bi[/latex] is [latex]a-bi[/latex]. 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.

- [latex]2+i\sqrt{5}[/latex]
- [latex]-\frac{1}{2}i[/latex]

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

- Write the division problem as a fraction.
- Determine the complex conjugate of the denominator.
- Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
- Simplify.

### Example: Dividing Complex Numbers

Divide [latex]\left(2+5i\right)[/latex] by [latex]\left(4-i\right)[/latex].

### Try It

### tip for success

We have seen that we can evaluate functions for any real number or algebraic expression we choose. For example,

given the function [latex]f(x)=2x-7[/latex],

[latex]f(-1) = 2(-1) -7 = -9[/latex]

[latex]f(5+h)=2(5+h)-7 = 10 +2h - 7 = 3 +2h[/latex].

We may also evaluate a function for a complex number if we wish, as shown in the Example and Try It boxes below.

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

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

### Try It

Let [latex]f\left(x\right)=2{x}^{2}-3x[/latex]. Evaluate [latex]f\left(8-i\right)[/latex].

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

Let [latex]f\left(x\right)=\dfrac{2+x}{x+3}[/latex]. Evaluate [latex]f\left(10i\right)[/latex].

### Try It

Let [latex]f\left(x\right)=\dfrac{x+1}{x - 4}[/latex]. Evaluate [latex]f\left(-i\right)[/latex].

## Simplifying Powers of [latex]i[/latex]

The powers of [latex]i[/latex] are cyclic. Let’s look at what happens when we raise [latex]i[/latex] to increasing powers.

[latex]{i}^{1}=i[/latex]

[latex]{i}^{2}=-1[/latex]

[latex]{i}^{3}={i}^{2}\cdot i=-1\cdot i=-i[/latex]

[latex]{i}^{4}={i}^{3}\cdot i=-i\cdot i=-{i}^{2}=-\left(-1\right)=1[/latex]

[latex]{i}^{5}={i}^{4}\cdot i=1\cdot i=i[/latex]

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

[latex]{i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1[/latex]

[latex]{i}^{7}={i}^{6}\cdot i={i}^{2}\cdot i={i}^{3}=-i[/latex]

[latex]{i}^{8}={i}^{7}\cdot i={i}^{3}\cdot i={i}^{4}=1[/latex]

[latex]{i}^{9}={i}^{8}\cdot i={i}^{4}\cdot i={i}^{5}=i[/latex]

### recall properties of exponents

Recall that [latex]\left(a^m\right)^n=a^{mn}[/latex].

### Example: Simplifying Powers of [latex]i[/latex]

Evaluate [latex]{i}^{35}[/latex].

### Q & A

**Can we write [latex]{i}^{35}[/latex] in other helpful ways?**

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

Factorization of [latex]{i}^{35}[/latex] |
[latex]{i}^{34}\cdot i[/latex] | [latex]{i}^{33}\cdot {i}^{2}[/latex] | [latex]{i}^{31}\cdot {i}^{4}[/latex] | [latex]{i}^{19}\cdot {i}^{16}[/latex] |

Reduced form |
[latex]{\left({i}^{2}\right)}^{17}\cdot i[/latex] | [latex]{i}^{33}\cdot \left(-1\right)[/latex] | [latex]{i}^{31}\cdot 1[/latex] | [latex]{i}^{19}\cdot {\left({i}^{4}\right)}^{4}[/latex] |

Simplified form |
[latex]{\left(-1\right)}^{17}\cdot i[/latex] | [latex]-{i}^{33}[/latex] | [latex]{i}^{31}[/latex] | [latex]{i}^{19}[/latex] |

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