complex numbers

Ans 4)

Put\ Z = r \left(\cos \theta + i\ \sin \theta \right) \\ Thus\ \left|Z \right| = r \\\\ Thus\ \frac{Z}{\left|Z \right|} = \cos \theta + i\ \sin \theta \\\ Hence\ \frac{Z}{\left|Z \right|}\ is\ unimodular

\left|\frac{Z}{|Z|}-\ 1 \right| = | \cos \theta + i\ \sin \theta -1| = 2 |\sin \theta /2| \leq 2 (\theta /2) = \theta = arg\ Z

Thus, Proved

1) is valid under the assumption that -\frac{\pi}{2} \le \arg z \le \frac{\pi}{2}

Then S = z^2 + \overline z = z^2 + \frac{1}{z} (\because \ |z| = 1 \Rightarrow z \overline z=1)

= \sqrt z \left(z^{\frac{3}{2}} + \frac{1}{z^{\frac{3}{2}}} \right)

But if |z| =1, \frac{1}{z^{\frac{3}{2}}} = \overline {z^{\frac{3}{2}}} \Rightarrow z^{\frac{3}{2}} + \frac{1}{z^{\frac{3}{2}}} is a real number.

If further we have the required constraints on arg z, then it is a positive real number.

Hence, then \arg (S) = \arg \sqrt z= \frac{1}{2} \arg z

2) |z| = 1 \Rightarrow z =\cos \theta + i \sin \theta

Let t = \tan \frac{\theta}{2}

Then z = \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2} i = \frac{1-t^2+2ti}{1+t^2} = \frac{(1-ti)^2}{(1+ti)(1-ti)}

=\frac{1-ti}{1+ti}

Setting c = \frac{1}{t} gives us the required result

thanks all of u for ur posts. by the way i got my answers himself but in a different way.