Question

1-2i/1-(1-i) ^2 convert into polar form ​

Answer 1

Step-by-step explanation:

Question is,

$z = \frac{1 - 2i}{1 - {(1 - i)}^{2} } \\ to \: find \: the \: polar \: form \: it \: is \: necessary \: to \: convert \: it \: into \: standerd \: form \\ z = \frac{1 - 2i}{1 - (1 + {i}^{2} - 2i)} \\ z = \frac{1 - 2i}{1 + 2i} \\ on \: rationalising \\ z = \frac{1 - 2i}{1 + 2i} \times \frac{1 - 2i}{1 - 2i} \\and \: on \: solving \\ z = \frac{ - 3}{5} - \frac{4i}{5}$

from here we can note that,

$x = \frac{ - 3}{5} \: and \: y = \frac{ - 4}{5}$

then,

${ |z| }^{2} = {r}^{2} = {( \frac{ - 3}{5} )}^{2} + {( \frac{ - 4}{5} )}^{2} \\ so \: |z| = r = 1$

we know that,

$\tan( \alpha ) = | \frac{imaginary}{real} | \\ on \: putting \: value \\ \tan( \alpha ) = | \frac{ \frac{ - 4}{5} }{ \frac{ - 3}{5} } | \\ \tan( \alpha ) = \frac{4}{3} \\ also \: \sin( \alpha ) = \frac{4}{5} \\ \: \: \: \ \: \: \: \: \: cos( \alpha ) = \frac{3}{5}$

on putting these value in the form,

$z = r(\cos( \alpha ) + i \sin( \alpha ) ) \\ \: z = 1( \ \ - cos( { \tan }^{ - 1} \frac{4}{3} ) + ( - i \sin( { \tan }^{ - 1} \frac{4}{3} )) \\ finally \\ z = ( - \cos( { \tan }^{ - 1} \frac{4}{3} ) - i \sin( { \tan }^{ - 1} \frac{4}{3} )$

thank you