calculating value for tangent in complex polar form. Polar form $z=re^{i\varphi}$, in which $\varphi \in (-\pi, \pi)$. Now $z$ is defined as:
$$z=\frac{4i-2}{5i+4}$$
we know that $$r=|z|=\frac{\sqrt{21}}{\sqrt{41}}$$
Now i would like to calculate value for $\tan \varphi$.
$$\tan \varphi = \frac{r \sin \varphi}{r \cos\varphi}$$
I dont know how to calculate value for $\tan \varphi$ since $\varphi$ is unknown. 
Any help would be much appreciated
Thanks,
Tuki
 A: First, we have for $z=x+iy$, with $x,y\in \mathbb{R}$
$$|z|=\sqrt{x^2+y^2}$$
and 
$$\arg(z)=\text{atan2}(y,x)$$
where $\text{atan2}(y,x)$ is defined HERE.

Therefore, we see that
$$\frac{-2+i4}{4+i5}=\frac{(-2+i4)(4-i5)}{41}=\frac{12}{41}+i\frac{26}{41}$$
such that 
$$\left|\frac{-2+i4}{4+i5}\right|=2\sqrt{\frac{5}{41}}$$
and 
$$\arg\left(\frac{-2+i4}{4+i5}\right)=\arctan(13/6)$$
A: You just have to compute first $\sin\varphi$ and $\cos\varphi$. 
Note that
$$z=\frac{4i-2}{5i+4}=\frac{(4i-2)(-5i+4)}{25+16}=\frac{12+26i}41=r(\cos\varphi+i\sin\varphi),$$
so we get the system of equations
$$\begin{cases}r\cos\varphi=\dfrac{12}{41},\\[1ex]r\sin\varphi=\dfrac{26}{41}\end{cases}\implies\begin{cases}r^2=\dfrac{12^2+26^2}{41^2},\\[1ex]\tan\varphi=\dfrac{r\cos\varphi}{\sin\varphi}=\dfrac{26}{12}=\dfrac{13}{6}\end{cases}\implies\begin{cases}r=\dfrac{2\sqrt{205}}{41},\\[1ex]\varphi\equiv\arctan\dfrac{13}{6}\mod\pi\end{cases}$$
As both $\sin \varphi$ and $\cos\varphi$ are positive and $-\pi<\varphi \le\pi$, we conclude  that 
$$\varphi=\arctan\dfrac{13}{6}\approx 1.138$$
