(d)$\frac{{1 + i}}{{1 – i}} = \frac{{1 + i}}{{1 – i}} \times \frac{{1 + i}}{{1 + i}} = \frac{{{{(1 + i)}^2}}}{2}$
Now $1 + i = r(\cos \theta + i\sin \theta ) \Rightarrow r\cos \theta = 1,r\sin \theta = 1$
==> $r = \sqrt 2 ,\theta = \pi /4$
$\therefore $ $1 + i = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)$
$ \Rightarrow \,$ $\frac{1}{2}\,{(1 + i)^2} = \frac{1}{2}\,.\,2\,{\left( {\cos \frac{\pi }{4} + i\,\sin \frac{\pi }{4}} \right)^2}$
By De Moivre's Theorem, $\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)$
Hence the amplitude is $\frac{\pi }{2}$ and modulus is $1$.
Trick : $arg{\rm{ }}\left( {\frac{{1 + i}}{{1 – i}}} \right) = arg(1 + i) – arg(1 – i)$
$ = {45^o} – ( – {45^o}) = {90^o}$
$\left| {\,\frac{{1 + i}}{{1 – i}}\,} \right|\, = \frac{{\left| {\,1 + i\,} \right|}}{{\left| {\,1 – i\,} \right|}}\, = \frac{{\sqrt 2 }}{{\sqrt 2 }} = 1$.