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 $k$?

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 $k = 4$.
