## Complex Analysis 1: Introduction

I wrote this because I thought I was teaching complex analysis, but then I switched so I might as well post this now.
Complex Analysis is the study of of the complex numbers (imaginary numbers). Most people are at least familiar with the number,
$i = \sqrt{-1}$

It came about in order to write a solution to the equation $x^2+1=0$ because no real number could solve it. Allowing for this “number” to exist gave rise to a lot of unforeseen consequences, which eventually developed into complex analysis.

If we have the number i, why can’t we have something like $3i$? We can. We can also add
$2i + 3i =5i$ just like we would with a real number.
$3 + i = ?$

We cannot simplify this any further like if we had two terms with $i$ in it. We just leave it as $3 + i$. Eventually we want to understand the significance of adding a real number with an imaginary number.

First, we say that a complex number is a number $a + bi$ where $a$ and $b$ are real numbers. $a$ is called the real part of the complex number, while $b$ is called the imaginary part of the complex number. Two complex numbers, $a+bi$ and $c+di$ are equal when $a=c \text{ and } b=d$. The following things are all complex numbers:

(1) $1 + i$
The real part is 1 and the imaginary part is also 1.

(2) $2 + 5i$
The real part is 2 and the imaginary part is 5.

(3) $0 + 3i = 3i$
The real part is 0 and the imaginary part is 3.

(4) $3 + 0i = 3$
The real part is 3 and the imaginary part is 0.

(5) $\frac{2}{9} - 12i$
The real part is $\frac{2}{9}$ and the imaginary part is -12.

(6) $\frac{29}{103} - \pi i$
The real part is $\frac{29}{103}$ and the imaginary part is $-\pi$.

From example (4) you can see one important thing. The real numbers are inside of the complex numbers, that is, every real number is also a complex number with a secret $+ 0i$ not written.

So we have complex numbers defined, we now want to see how to put them together in various ways.
Addition of two complex numbers $a+bi, c+di$.

$(a + bi) + (c + di) = (a+c) + (bi+di) = (a+c) + (b+d)i$

You just add the real parts together, and then add the imaginary parts together. For example,

$(2 - 3i) + (5 + 2i) = (2+5) + (-3i+2i) = 7 - 1i$

Subtraction is roughly the same.
$(a + bi) - (c + di) = (a-c) + (bi-di) = (a-c) + (b-d)i$

We still put the real parts together and then the imaginary parts together, while making sure to keep the negatives where they should be.

Next is multiplication. For this, we foil out just like we would with anything else. For example
$(3+5)(2+7) = (3)(2) + (3)(7) + (5)(2) + (5)(7) = 6+21+10+35 = 72$

$(a + bi)(c + di) = (a)(c) + (a)(di)+(bi)(c)+(bi)(di) = ac + adi + bci + bdi^2$
Remember that
$i = \sqrt{-1}$
so that
$i^2 = -1$

Doing this and combining the imaginary parts we get
$ac + adi + bci - bd$
$(ac - bd) + adi + bci$
$(ac - bd) + (ad + bc)i$
An example would be
$(2+i)(3-2i) = 6-2i+3i-2i^2 = 6-2i+3i+2= 6+2-2i+3i = 8+i$

Lastly is division:
$\frac{a+bi}{c+di}$

You may remember rationalizing the denominator for when you had square roots on the bottom of a fraction. This is similar, we real-ize the denominator instead. (Multiply by 1 in a sly way)

$\frac{a+bi}{c+di}= (\frac{a+bi}{c+di})(\frac{c-di}{c-di})$

$= \frac{(a+bi)(c-di)}{(c+di)(c-di)}$

Using how multiplication is defined we get:

$= \frac{(ac+bd) + (-ad+bc)i}{(c^2+d^2 -cdi+cdi)}$

Here you can see where the imaginary part cancels out to zero on the denominator so we get:

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

An example:

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

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

$= \frac{(1-12) + (3+4)i}{(1^2+(-4)^2 -4i+4i)}$

$= \frac{(-11 + 7i)}{1+16}$

$= \frac{(-11 + 7i)}{1+16}$

$= \frac{(-11 + 7i)}{17}$

Which we can leave like this or write:

$= \frac{-11}{17} + \frac{7i}{17}$

You don’t have to worry too much about multiplication and division in this way because there is a much easier way in the future, but one important thing happened with division

Notice that
$(a+bi)(a-bi) = (a)(a) +abi - abi - (b)(b)i^2$
$= a^2 +b^2$

When given a complex number $a+bi$ the number $a-bi$ is called the complex conjugate of $a+bi$. In general you just change the sign of the imaginary part.
For example:

The complex conjugate of $3+i$ is $3-i$. That’s the easy part. The nice part is what I did earlier, when you multiply a number by its conjugate you get $a^2 + b^2$

Next time I will show another way of representing complex numbers.