# 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 $\frac{1}{100}$
Hint: let $x = \alpha - \beta$. Then
$$\tan^2 \theta = \frac {\sin^2(x)}{(\cos (x) -10)^2}:=g(x).$$
The function $g$ has local extrema at $\cos(x) =1/10$ and $\sin(x) = 0$.
Evaluate $g$ at the corresponding $x$ values.
Remark Don't forget to test for absolute Extrema by checking the endpoints of the correct domain for $x$.