Solving for $k$ when $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$ Consider $$|z|=|z-3i|$$ 
We know that if $z=a+bi\Rightarrow b=\frac{3}{2}$
$z_1$ and $z_2$ will represent two possible values of $z$ such that $|z|=3$. We are given $\arg(z_1)=\frac{\pi}{6}$
The value of $k$ must be found assuming $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$ 
My attempt:
We know $z_1=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$ and $z_2=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$ by solving for $a$.
So let $z_3=\frac{z_1^kz_2}{2i}$ $$z_3=\frac{z_1^kz_2}{2i} = \frac{i\left(\frac{3\sqrt{3}}{2}+\frac{3}{2}i\right)^k\left(-\frac{3\sqrt{3}}{2}+\frac{3}{2}i\right)}{-2}$$
We know $$\arg(z_3)=\arctan\left(\frac{\operatorname{Im}(z_3)}{\operatorname{Re}(z_3)}\right)=\pi \Rightarrow \frac{\operatorname{Im}(z_3)}{Re(z_3)}=\tan(\pi)=0$$
This is the part where I get stuck; I assume that $\operatorname{Re}(z_3)\neq0$ and then make the equation $$\operatorname{Im}(z_3)=0$$
However, I am not sure on how to get the value of $k$ from this, or if I am in the right direction.
What should I do in order to get the value of $k$?
 A: Hint:
$$
|z|=|z-3i|
$$
means that the distances of $z$ from $0$ and $3i$ are the same, so $z$ stay on the axis of the segment......
A: $$z_1=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$ 
and 
$$z_2=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
Thus,
$$z_1=3e^{\frac{i\pi}{6}}$$ and $$z_2=3e^{\frac{5i\pi}{6}}$$
$$i=e^{\frac{i\pi}{2}}$$
$$\arg\left(\frac{z_1^kz_2}{2i}\right)=\arg\left(\frac{3^{k+1}}{2}\frac{e^{\frac{ki\pi}{6}}e^{\frac{5i\pi}{6}}}{e^{\frac{i\pi}{2}}}\right)=(k+2)\frac{\pi}{6}$$
$$(k+2)\frac{\pi}{6}=\pi$$
$$k=4$$
A: You already know that $\arg(z_1)=\frac{\pi}{6}$, and moreover $\arg(z_2)=\frac{5\pi}{6}$ and $\arg \left(\frac{1}{2i}\right)=\frac{-\pi}{2}$. Multiplying complex numbers results in adding their arguments (modulo $2\pi$) so you get the equation
$$\arg\left(\frac{z_1^kz_2}{2i}\right)=k\frac{\pi}{6}+\frac{5\pi}{6}-\frac{\pi}{2}=\frac{(k+2)\pi}{6}=\pi+2m\pi$$which gives you$$k=4+12m$$where $m\in \mathbb{Z}$.
A: I would go for the polar form of $z_1$ and $z_2$. I do this because exponentiation of complex numbers is much easier with polar forms. I also assume that your argument is contained in $[0,2\pi)$, so
$z_1 = 3e^{i\pi/6}$ and $z_2 = 3e^{i5\pi/6}$, and then
$$z_1^kz_2 = 3e^{i\pi k/6}3e^{i5\pi/6} = 9e^{i\pi/6(k+5)},$$
and since $2i = 2e^{i\pi/2}$, we have also
$$\frac{z_1^kz_2}{2i} = \frac{9}{2}\frac{e^{i\pi/6(k+5)}}{e^{i\pi/2}} = \frac{9}{2}e^{i\pi/6(k+5-3)}=\frac{9}{2}e^{i\pi/6(k+2)}.$$
You want the argument to be equal to $\pi$, so
$$\pi/6(k+2) = \pi,$$
which gives $k = 4$.
