If $z=a+bi$, what is the relationship between $z$ and $zi$? 
If $z=a+bi$, what is the relationship between $z$ and $zi$?

My efforts:
$$\begin{align}
z\phantom{i}  &= \phantom{-}a + bi \\
zi &=-b + ai
\end{align}$$
If I write these points in a coordinate system I will see there is a $90^\circ$ angle between them. If I continue, then I'll get a $90^\circ$ angle again between the two points.
Since the argument for $z$ is
$$\tan\theta = \frac{b}a \tag1$$
$$\tan\theta_1 = \frac{-a}{b} \tag2$$
$$\theta_1 -\theta = \frac{\pi}{2} \tag3$$
Therefore,
$$\arctan{\frac{-a}b} - \arctan{\frac{b}a} = \frac{\pi}2 \tag{$\star$}$$
How would one show that the last equation is true?
 A: You might know that
for any real $ x>0$, we have
$$\arctan(x)+\arctan\left(\frac 1x\right)=\frac{\pi}{2}$$
and
and if $ x<0$,
$$\arctan(x)+\arctan\left(\frac 1x\right)=\frac{-\pi}{2}$$
what about the case $\;x=\dfrac ab\;$, using the fact that $ \arctan $ is an odd function.
A: I am summarizing all the comments.
(1)
Let $ z= a+ ib.$ Then $iz= -b +ia.$ Note $ z= a+ ib$ reprsent the vector $ (a, b) \in \mathbb{R^2}.$ Then observe that  the dot/inner product of these two vectors: $  (a, b). (-b, a) = -ab +ab =0.$ This shows this two vectors are orthogonal/perpendicular to each other. This proves $arg(z)  \sim arg (iz) = \frac{\pi}{2}.$
(2) Recall that the geometric interpretation of complex number multiplication is the following: Let $z$ be a given a complex number. If we multiply $z$ by another complex number $w,$ then $w$ rotates $z$ by angle $\theta$ (this angle is the argument of w) and at the same time $w$ stretches/compresses $z$ by a factor $|w|$ (the norm/modulus of $w$).
Since $i$ has norm/modulus $1,$ it does not change the length/norm/modulus of any complex number $z$ when we multiply by $i.$ Surely it rotates any complex number by angle $\pi/2$ because $i$ has argument $\pi/2.$ This shows that $arg(z)  \sim arg (iz) = \frac{\pi}{2}.$
(3) What I have written above can be explained more efficiently by the use of the polar form of a complex number. Any complex number $z$ can be expressed as $re^{i \theta},$ $r$ is the modulus of $z$ and $\theta$ is the argument of $z.$ If we have two complex numbers $z$ and $w,$ then $zw= |z| |w| e^{i(\theta +\beta)},$ where $\theta$ and $\beta$ are the arguments of $z$ and $w$ respectively. Using this formula, we have that if $ z= |z|e^{i \theta}$, then $iz = 1. |z|e^{i (\theta + \pi/2)} = |z|e^{i (\theta + \pi/2)}$. This proves that $arg(z)  \sim arg (iz) = \frac{\pi}{2}.$
A last warning: the angle $\theta$ associated with the complex number $z$ is called its argument, NOT $\tan(\theta).$ I believe you know the right definition.
