Compute all possible values of $\cot{\theta} - \frac{6}{z}$. Note that $ 0 < \theta < \frac{\pi}{2}$. 
Let $\theta = \arg{z}$ and suppose $z$ satisfies $|z - 3i| = 3$.
Compute all possible values of $\displaystyle \cot{\theta} - \frac{6}{z}$. Note that $\displaystyle 0 < \theta < \frac{\pi}{2}$.

$\bf{My\; Try::}$ Let $z-3i=3e^{i\theta}\Rightarrow z=3i+3\cos \theta+3i\sin \theta = 3\cos \theta+3i(1+\sin \theta)$
So we get $$z=3\sin \left(\frac{\pi}{2}-\theta\right)+3i\left[1+\cos\left(\frac{\pi}{2}-\theta\right)\right]$$
So we get $$z=3i\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)e^{-i\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$$
Now how can i solve after that, Thanks
 A: Since $\theta=\arg z$, note that $z=3\cos\theta+3i(1+\sin\theta)$ is not correct.

Let $z=x+yi$ where $x,y\in\mathbb R$. Then, since we have
$$x^2+(y-3)^2=3^2$$we can write
$$x=3\cos\alpha,\qquad y=3+3\sin\alpha$$
where $-\pi/2\lt\alpha\lt \pi/2$.
So, we can have
$$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}=\frac{3\cos\alpha}{\sqrt{6(3+3\sin\alpha)}}$$
$$\sin\theta=\frac{y}{\sqrt{x^2+y^2}}=\frac{3+3\sin\alpha}{\sqrt{6(3+3\sin\alpha)}}$$
and so
$$\begin{align}\cot\theta-\frac{6}{z}&=\frac{3\cos\alpha}{3+3\sin\alpha}-\frac{6}{3\cos\alpha+(3+3\sin\alpha)i}\\&=\frac{\cos\alpha}{1+\sin\alpha}-\frac{2(\cos\alpha-(1+\sin\alpha)i)}{\cos^2\alpha+(1+\sin\alpha)^2}\\&=\frac{\cos\alpha}{1+\sin\alpha}-\frac{\cos\alpha-(1+\sin\alpha)i}{1+\sin\alpha}\\&=\frac{1+\sin\alpha}{1+\sin\alpha}i\\&=\color{red}{i}\end{align}$$
A: WLOG $z-3i=3(\cos2t+i\sin2t)$
$\iff z=3\{\cos2t+i(1+\sin2t)\}$
$=6\sin\left(\dfrac\pi4-t\right)\left[\cos\left(\dfrac\pi4-t\right)+i\sin\left(\dfrac\pi4-t\right)\right]$
$\dfrac6z=\cdots=\dfrac{\cos\left(\dfrac\pi4-t\right)-i\sin\left(\dfrac\pi4-t\right)}{\sin\left(\dfrac\pi4-t\right)}=\cot\left(\dfrac\pi4-t\right)-i$
$\tan\theta=\cdots=\tan\left(\dfrac\pi4-t\right)$
Can you take it from here?
