By the end of this section, you should be able to:

1. Describe complex numbers

2. Solve problems involving complex numbers

This one's *really short*. Let's dive right in.

A complex number is an expression in the form \(a+bi\), where \(a\) represents the real number part, and \(bi\) represents the imaginary part, where \(i^2 = -1\).

Every complex number has its conjugate, which has the opposite sign between the real and imaginary terms: \(a-bi\). Some textbooks represent the conjugate of some expression \(e = a+bi\), as \(\overline{e}=a-bi\), using an overline.

We can treat complex numbers just as we would polynomials. I'll prove it to you with a few examples:

Simplify \((4+3i) + (5+2i)\)\(\)

Group the like terms:

\((4+5) + i(3+2)\)

\(9 + 5i\).

Now something a bit tricker.

If \(z = 6+3i\), determine \(z\overline{z}\)

\((6+3i)(6-3i)\)

\(=36 - i^2(9)\)

\(=36-(-1)(9)\)

\(=45\)

Even trickier!

Write \(\frac{1}{3-2i}\)in \(a+bi\) form.

In this case, we need to get rid of the imaginary number from the denominator. How do we do that? Well, we multiply the numerator and denominator by the conjugate of the denominator. That's a lot of words! Let me show you:

\(=\frac{1}{3-2i} \times \frac{3+2i}{3+2i}\)

\(=\frac{3+2i}{(3-2i)(3+2i)}\)

\(=\frac{3+2i}{9-4i^2}\)

\(=\frac{3+2i}{9-4(-1)}\)

\(=\frac{3+2i}{13}\)

There we go! It's in \(a+bi\) form.

That's it. Told you it was short! Don't forget to try out the practice problems.

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