Let z be a complex number (not lying on X-axi | vedclass

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Question

(b)Let $z = r(\cos 	heta + i\sin 	heta )$.
Then $\left| {z + \frac{1}{z}} ight| = 1\,\, \Rightarrow {\left| {z + \frac{1}{z}} ight|^2}$=1
==&g

Vedclass · Accepted Answer

${\mathop{m Re}
olimits} (z) = 0$

Vedclass · Answer

(b)Let $z = r(\cos 	heta + i\sin 	heta )$.
Then $\left| {z + \frac{1}{z}} ight| = 1\,\, \Rightarrow {\left| {z + \frac{1}{z}} ight|^2}$=1
==> ${\left| {r(\cos 	heta + i\sin 	heta ) + \frac{1}{r}(\cos 	heta - i\sin 	heta )} ight|^2} = 1$.
==> ${\left( {r + \frac{1}{r}} ight)^2}{\cos ^2}	heta + {\left( {r - \frac{1}{r}} ight)^2}{\sin ^2}	heta = 1$
==> ${r^2} + \frac{1}{{{r^2}}} + 2\cos 2	heta = 1$
Since $|z| = r$ is maximum, therefore $\frac{{dr}}{{d	heta }} = 0$
Differentiating $(i)$ w.r.t.$	heta $, we get
$2r\frac{{dr}}{{d	heta }} - \frac{2}{{{r^3}}}\,\frac{{dr}}{{d	heta }} - 4\sin 2	heta = 0$
Putting $\frac{{dr}}{{d	heta }} = 0$,we get $\sin 2	heta = 0$==> $	heta = 0$or $\frac{\pi }{2}$
==> $z$ is purely imaginary or purely real.

Let $z$ be a complex number (not lying on $X$-axis) of maximum modulus such that $\left| {z + \frac{1}{z}} \right| = 1$. Then

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