I am trying to prove that for any $z \in \mathbb{C}$, we can write $z = re^{ i \theta}$ where $r = |z|$ and $\theta$ is the polar angle. I do not understand the solution that I am looking at, and was hoping someone could look over my understanding of the argument.
The first step is to write: \begin{align*} re^{i \theta} & = |z|e^{ i \theta} & & \text{given $r = |z|$} \\ & = |z|\left(\cos \theta + i \sin \theta\right) & & \text{Euler's identity} \\ & = |z| \cos \theta + (|z| \sin \theta) i \end{align*} So far I haven't any issue with anything in this solution. The next step, however, does not make sense to me. The assertion is that "by trigonometry, we have" $$z = |z|\cos \theta + (|z| \sin \theta)i.$$ And equality is obvious from this point by equating this with $re^{i \theta}$ above. I do not understand how to derive this fact using trigonometry. The best I can do with trigonometry is, for some arbitrary point in the complex plane with radius $r$ and polar angle $\theta$, draw a right triangle with sides $x = r \cos \theta$, $y = r \sin \theta$, and $r = x^2 + y^2$.
Any help with this would be appreciated.
You know that any complex number $z$ is written as $$ z=a+bi,\quad a,b\in\mathbb R. $$ You have to prove that this can be always written in the form: $$ z=r e^{i\theta}=r(\cos\theta+i\sin\theta). $$ So, consider $z=a+bi$. If $z=0$, then you have $z=0\cdot e^{i\theta}$ for any $\theta$ and you have finished. If $z\neq 0$, define $$ r=\sqrt{a^2+b^2}>0 $$ and write $$ z=a+ib=r\left(\frac {a}{r}+\frac {b}{r}\right). $$ Then $ \frac {a}{r}\in [-1,1]$ and you can define $$ \theta=\pm\arccos(a/r)\Rightarrow a/r=\cos\theta. $$ Then you prove that, choosing the appropriate sign, $\sin\theta=b/r. $ Indeed $$ \left(\frac{a}{r}\right)^2+\left(\frac{b}{r}\right)^2=1\Rightarrow \frac{b}{r}=\pm\sqrt{1-(a/r)^2}=\pm\sqrt{1-\cos^2\theta} $$