Argument and modulus of frac{{1 + i}}{{1 - i} | vedclass

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Question

(d)$\frac{{1 + i}}{{1 - i}} = \frac{{1 + i}}{{1 - i}} 	imes \frac{{1 + i}}{{1 + i}} = \frac{{{{(1 + i)}^2}}}{2}$
Now $1 + i = r(\cos 	heta + i\sin

Vedclass · Accepted Answer

$\frac{\pi }{2}$and $1$

Vedclass · Answer

(d)$\frac{{1 + i}}{{1 - i}} = \frac{{1 + i}}{{1 - i}} 	imes \frac{{1 + i}}{{1 + i}} = \frac{{{{(1 + i)}^2}}}{2}$
Now $1 + i = r(\cos 	heta + i\sin 	heta ) \Rightarrow r\cos 	heta = 1,r\sin 	heta = 1$
==> $r = \sqrt 2 ,	heta = \pi /4$
$	herefore $ $1 + i = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} ight)$
$ \Rightarrow \,$ $\frac{1}{2}\,{(1 + i)^2} = \frac{1}{2}\,.\,2\,{\left( {\cos \frac{\pi }{4} + i\,\sin \frac{\pi }{4}} ight)^2}$
By De Moivre's Theorem, $\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} ight)$
Hence the amplitude is $\frac{\pi }{2}$ and modulus is $1$.
Trick : $arg{m{ }}\left( {\frac{{1 + i}}{{1 - i}}} ight) = arg(1 + i) - arg(1 - i)$
$ = {45^o} - ( - {45^o}) = {90^o}$
$\left| {\,\frac{{1 + i}}{{1 - i}}\,} ight|\, = \frac{{\left| {\,1 + i\,} ight|}}{{\left| {\,1 - i\,} ight|}}\, = \frac{{\sqrt 2 }}{{\sqrt 2 }} = 1$.

List-$I$	List-$II$
($P$) $\|z\|^2$ is equal to	($1$) $12$
($Q$) $\|z-\bar{z}\|^2$ is equal to	($2$) $4$
($R$) $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	($3$) $8$
($S$) $\|z+1\|^2$ is equal to	($4$) $10$
	($5$) $7$

Argument and modulus of $\frac{{1 + i}}{{1 - i}}$ are respectively

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