If $z$ and $\omega$ are two complex no. and $\theta = \arg\left(\frac{\omega -z}{z}\right)\;,$ Then Max. of $\tan^2 \theta$ 
If $\omega$ and $z$ are two complex number such that $|\omega| = 1$ and $|z|=10$ and Let $\displaystyle \theta = \arg\left(\frac{\omega -z}{z}\right)$
Then Maximum possible value of $\tan^2 \theta$

$\bf{My\; Try::}$ Let $\omega = =e^{i\alpha} = \cos \alpha+i\sin \alpha$ and $z = 10e^{i\beta}=10(\cos \beta+i\sin \beta)$
Given $\displaystyle \theta = \arg\left(\frac{\omega -z}{z}\right)$
Now $$\displaystyle \frac{\omega-z}{z} = \frac{\omega}{z}-1 = \frac{1}{10}e^{i(\alpha-\beta)}-1 = \frac{1}{10}\left[\cos (\alpha-\beta) +i\sin (\alpha-\beta) \right]-1$$
So we get $$\displaystyle \frac{\omega}{z}-1 = \frac{1}{10}\cos (\alpha -\beta)-1+i\frac{1}{10}\sin (\alpha-\beta)$$
So $$\displaystyle \tan \theta = \frac{\sin (\alpha-\beta)}{\cos(\alpha - \beta)-10}$$

$f(x)=\frac{sinx}{cosx-10}$ is $2\pi$ periodic and is decreasing in first quadrant and fourth quadrants and increasing in second and third. so obviously max of $tan^2 \theta$ occurs at $x=\frac{\pi}{2}$ Or $x=\frac{3\pi}{2}$ so max is $\frac{1}{100}$
