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Commit 1b72e489 authored by Timo1104's avatar Timo1104
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Added division example

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1 merge request!2Add lecture on complex numbers properly
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......@@ -59,6 +59,12 @@ complex conjugate of $z_2$:
$$\frac{z_1}{z_2} = \frac{z_1 z_2^*}{z_2 z_2^*} = \frac{(a_1 a_2 + b_1 b_2) + (-a_1 b_2 + a_2 b_1) {{\rm i}}}{a_2^2 + b_2^2}.$$
Check this!
**Example**
$$\begin{align}
\frac{1 + 2{\rm i}}{1 - 2{\rm i}} & = (\frac{(1 + 2{\rm i})(1 + 2{\rm i})}{1^2 + 2^2}) &= \frac{1+8{\rm i} -4}{5}\\
&= -\frac{3}{5} + {\rm i} \frac{8}{5}
\end{align}$$
## The complex plane
Complex numbers can be rendered on a two-dimensional (2D) plane, the
......@@ -115,8 +121,11 @@ $$\varphi = \begin{cases} \arctan(b/a) &{\rm for ~} a>0; \\
It turns out that using this magnitude $|z|$ and phase $\varphi$, we can write any complex number as
$$z = |z| e^{{\rm i} \varphi}$$
When increasing $\varphi$ with $2 \pi$, we reach the same point on the complex plane. In other words, when adding $2 \pi$ to our argument, we get the same complex number!
As a result, the argument is defined up to $2 \pi$, and we are free to make any choice we like:
When increasing $\varphi$ with $2 \pi$, we make a full circle and reach the same point on the complex plane. In other words, when adding $2 \pi$ to our argument, we get the same complex number!
As a result, the argument $\varphi$ is defined up to $2 \pi$, and we are free to make any choice we like:
$$\begin{align}
-\pi < \varphi < \pi $ {\rm (left image)} \\
-\frac{\pi}{2} < \varphi < \frac{3 \pi}{2} $ {\rm (right image)} \end{align} $$
![image](figures/complex_numbers_11_0.svg)
......@@ -124,11 +133,10 @@ As a result, the argument is defined up to $2 \pi$, and we are free to make any
Real functions can (most of the times) be written in terms of a Taylor series:
$$f(x) = \sum \limits_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n}$$
We can write something similar for complex functions,
when replacing $x \rightarrow z$:
when replacing the *real* variable $x$ with its *complex* counterpart $z$:
$$f(z) = \sum \limits_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (z-x_{0})^{n}$$
We have already seen that we can write any complex number $z$ in polar form as, $z = |z| e^{{\rm i} \varphi}$. Apparently, we can write any complex number using the complex
exponential function, at which we will have a look below.
For this course, the most important function is the *complex exponential function*, at which we will have a look below.
### The complex exponential function
The complex exponential is used *extremely often*.
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