Answer:

Question is,

\BEGIN{gathered}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} \\ \end{gathered}

z=

1−(1−i)

2

1−2i

tofindthepolarformitisnecessarytoconvertitintostanderdform

z=

1−(1+i

2

−2i)

1−2i

z=

1+2i

1−2i

onrationalising

z=

1+2i

1−2i

×

1−2i

1−2i

andonsolving

z=

5

−3

−

5

4i

from here we can note that,

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

5

−3

andy=

5

−4

then,

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

∣z∣

2

=r

2

=(

5

−3

)

2

+(

5

−4

)

2

so∣z∣=r=1

we know that,

\begin{gathered} \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} \end{gathered}

tan(α)=∣

real

imaginary

∣

onputtingvalue

tan(α)=∣

5

−3

5

−4

∣

tan(α)=

3

4

alsosin(α)=

5

4

cos(α)=

5

3

on putting these value in the form,

\begin{gathered}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} )\end{gathered}

z=r(cos(α)+isin(α))

z=1( −cos(tan

−1

3

4

)+(−isin(tan

−1

3

4

))

finally

z=(−cos(tan

−1

3

4

)−isin(tan

−1

3

4

)

thank you