$r\angle \theta = r(\cos \theta + i \sin \theta)$ So I am looking at the Visual Complex Analysis book and the first practice section asks to go through some of the statements about complex numbers and just prove to oneself that they are true. There is one that uses the notation in the title and that says that the angle between the $x$ axis and the complex number vector is $\cos \theta + i\sin \theta$. I know that this is something very simple and yet I cannot show this...
 A: Any complex number $z$ can be expressed in 3 equivalent ways:


*

*Algebraic form
$$z = a + ib,$$
for some $a \in \mathbb{R}$ and $b \in \mathbb{R}.$

*Trigonometric form
$$z = r(\cos(\theta) + i \sin(\theta)),$$


where
$$r = \sqrt{a^2 + b^2},$$
and $\theta$ is the angle between the x axis and the complex number vector. $\theta$ is called the argument of $z$. Please, read this for more details.


*Exponential form
$$z = re^{i\theta},$$
where $r$ and $\theta$ are defined as before.
A: Can you get something from this picture?


[Added in response to the comments] Note that in the notation $\angle\theta$, $\theta$ is the argument (angle) of $z$, but the whole thing $\angle\theta$ is not. Read page 7 of Needham's book about the property of $\angle\theta$. Note that, for any $\theta$ and $\phi$, we have $\angle(\theta+\phi)=\angle\theta+\angle\phi$. But it is not true that $\theta\phi=\theta+\phi$ (if $\angle\theta=\theta$, this would be true). Reading the later chapter of the book, you will find later that $\angle\theta=e^{i\theta}$.
