# How To Simplify Complex Numbers

Algebra often involves simplifying expressions, but some expressions are more confusing to deal with than others. Complex numbers involve the quantity known as *i*, an "imaginary" number with the property *i* = √−1. If you have to simply an expression involving a complex number, it might seem daunting, but it's quite a simple process once you learn the basic rules.

### TL;DR (Too Long; Didn't Read)

Simplify complex numbers by following the rules of algebra with complex numbers.

## What is a Complex Number?

Complex numbers are defined by their inclusion of the *i* term, which is the square root of minus one. In basic-level mathematics, square roots of negative numbers don't really exist, but they occasionally show up in algebra problems. The general form for a complex number shows their structure:

\(z = a + bi\)

Where *z* labels the complex number, *a* represents any number (called the "real" part), and *b* represents another number (called the "imaginary" part), both of which can be positive or negative. So an example complex number is:

\(z = 2 −4i\)

Since all square roots of negative numbers can be represented by multiples of *i*, this is the form for all complex numbers. Technically, a regular number just describes a special case of a complex number where *b* = 0, so all numbers could be considered complex.

## Basic Rules for Algebra with Complex Numbers

To add and subtract complex numbers, simply add or subtract the real and imaginary parts separately. So for complex numbers *z* = 2 – 4*i* and *w* = 3 + 5*i*, the sum is:

\(\begin{aligned}

z + w &= (2 – 4i) + (3 + 5i)\)

\(&=(2 + 3) + (-4 + 5)i\)

\(&= 5 + 1i \

&= 5 + i

\end{aligned}\)

Subtracting the numbers works in the same way:

\(\begin{aligned}

z- w &= (2 – 4i) – (3 + 5i) \

&= (2 – 3) + (-4 – 5)i \

&= -1 -9i

\end{aligned}\)

Multiplication is another simple operation with complex numbers, because it works like ordinary multiplication except you have to remember that *i*^{2} = −1. So to calculate 3*i* × −4*i*:

\(3i × -4i = -12i^2\)

But since *i*^{2}= −1, then:

\(-12i^2 = -12 ×-1 = 12\)

With full complex numbers (using *z* = 2 – 4*i* and *w* = 3 + 5*i* again), you multiply them in the same way you would with ordinary numbers like (*a* + *b*) (*c* + *d*), using the "first, inner, outer, last" (FOIL) method, to give (*a* + *b*) (*c* + *d*) = *ac* + *bc* + *ad* + *bd*. All you have to remember is to simplify any instances of *i*^{2}. So for example:

\(\begin{aligned}

z × w &= (2 -4i)(3 + 5i) \

&= (2 × 3) + (-4i × 3) + (2 × 5i) + (−4i × 5i) \

&= 6 -12i + 10i – 20i^2 \

&= 6 -2i + 20 \

&= 26 + 2i

\end{aligned}\)

## Dividing Complex Numbers

Dividing complex numbers involves multiplying the numerator and denominator of the fraction by the complex conjugate of the denominator. The complex conjugate just means the version of the complex number with the imaginary part reversed in sign. So for *z* = 2 – 4*i*, the complex conjugate *z** = 2 + 4*i*, and for *w* = 3 + 5*i*, *w** = 3 −5*i*. For the problem:

\(\frac{z}{w} = \frac{2 -4i}{3 + 5i}\)

The conjugate needed is *w**. Divide the numerator and denominator by this to give:

\(\frac{z}{w} = \frac{(2 -4i)(3 -5i)}{(3 + 5i)(3-5i)}\)

And then you work through as in the previous section. The numerator gives:

\(\begin{aligned}

(2 -4i) (3 -5i) &= 6 -12i- 10i + 20i^2 \

&= -14-22i

\end{aligned}\)

And the denominator gives:

\(\begin{aligned}

(3 + 5i)(3-5i) &= 9 + 15i – 15i -25i^2 \

&= 9 + 25 \

&= 34

\end{aligned}\)

This means:

\(\begin{aligned}

\frac{z}{w} &= \frac{-14 – 22i}{34} \

\,\

&= \frac{-14}{34} – \frac{22i}{34} \

\,\

&= \frac{-7}{17} -\frac{11i}{17}

\end{aligned}\)

## Simplifying Complex Numbers

Use the rules above as needed to simplify complex expressions. For example:

\(z = \frac{(4 + 2i) + (2 -i)}{(2 + 2i)(2+ i)}\)

This can be simplified by using the addition rule in the numerator, the multiplication rule in the denominator, and then completing the division. For the numerator:

\((4 + 2i) + (2 – i) = 6 + i\)

For the denominator:

\(\begin{aligned}

(2 + 2i)(2+ i) &= 4 + 4i + 2i + 2i^2 \

&= (4 -2) + 6i \

&= 2 + 6i

\end{aligned}\)

Putting these back in place gives:

\(z = \frac{6 + i}{2 + 6i}\)

Multiplying both parts by the conjugate of the denominator leads to:

\(\begin{aligned}

z &= \frac{(6 + i) (2 – 6i)}{(2 + 6i) (2 -6i)} \

\,\

&= \frac{12 + 2i -36i -6i^2}{4 + 12i -12i -36i^2} \

\,\

&= \frac{18 – 34i}{40} \

\,\

&= \frac{9 – 17i}{20} \

\,\

&= \frac{9}{20} -\frac{17i}{20} \

\end{aligned}\)

So this means *z* simplifies as follows:

\(\begin{aligned}

z &= \frac{(4 + 2i) + (2 – i)}{(2 + 2i)(2+ i)} \

&= \frac{9}{20} -\frac{17i}{20} \

\end{aligned}\)

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