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