Suppose z ∈ C has argument θ such that 0 < θ < π 2 and satisfy the equation | z…

Given,

$\left|z-3i\right|=3$ which represents a circle with radius as $3$ and centre as $(0,3)$

Now,

Z is lying anywhere on $\left|z-3i\right|=3$

From the figure below, we see that

$\angle OAB=90°$ (Angle subtended in a semi-circle)

So,

$$\begin{gathered} OA=Z \\ OB=6 \end{gathered}$$

$OA=OB \sin \theta e^{i\theta }$

$\Rightarrow Z=6 \sin \theta e^{i\theta }$

$\Rightarrow \frac{6}{Z}=\frac{1}{\sin \theta \left(e^{i\theta }\right)}$

$\Rightarrow \frac{6}{Z}=\frac{1}{\sin \theta (\cos \theta +i\sin \theta )}$

By rationalising, we get

$\Rightarrow \frac{6}{Z}=\frac{\cos \theta - i \sin \theta }{\sin \theta (\cos \theta +i\sin \theta )(\cos \theta -i\sin \theta )}$

$\Rightarrow \frac{6}{Z}=\frac{\cos \theta - i \sin \theta }{\sin \theta }$

$\Rightarrow \frac{6}{Z}=cot\theta -i$

$\Rightarrow cot\theta -\frac{6}{Z}=i$.