Question

find the modulus and argument of i/(1-i)​

Answer 1

$\implies \: \frac{ \iota}{(1 - \iota)} \\ \\ \implies \: \frac{ \iota(1 + \iota)}{(1 - \iota)(1 + \iota)} \\ \\ \implies \: \frac{ \iota + \iota {}^{2} }{1 - \iota {}^{2} } \\ \\ \implies \: \frac{ \iota - 1}{1 - ( - 1)} \\ \\ \implies \: \frac{ - 1 + \iota}{1 + 1} \\ \\ \implies \frac{ - 1 + \iota}{2} \\ \\ \implies \: - \frac{1}{2} + \frac{ \iota}{2} \\ \\ modulas \\ \\ \implies \: |z| = \sqrt{ (\frac{ \: \: \: \: 1 }{ - 2} ) {}^{2} + ( \frac{1}{2} ) {}^{2} } \\ \\ \implies \: |z| = \sqrt{ \frac{1}{4} + \frac{1}{4} } \\ \\ \implies \: |z| = \sqrt{ \frac{1 + 1}{4} } \\ \\ \implies \: |z| = \sqrt{ \frac{ \cancel{2}}{ \cancel{4} {}^{2} } } \\ \\ \implies \: |z| = \frac{ \: \: \: 1}{ \sqrt{2} } \\ \\ arguement \\ \\ \implies \tan {}^{ - 1} | \frac{y}{x} | \\ \\ \implies \tan {}^{ - 1} \cancel{ | \frac{ \frac{ \: \: \: 1}{ - 2} }{ \frac{1}{2} } | } \\ \\ \implies \tan {}^{ - 1} ( 1) \\ \\ \implies \cancel{ \tan {}^{ - 1} ( \tan }\: \frac{ \pi}{4} ) \\ \\ \implies \: \frac{ \pi}{4} \\ \\ ( - . + ) \:second \: quardent \\ \\ \implies( \pi - \theta) \\ \\ \implies \pi - \frac{ \pi}{4} \\ \\ \implies \: \frac{4 \pi - \pi}{4} \\ \\ \implies \: \frac{3 \pi}{4}$

_________________________

thanks