Question 1177965


{{{1/(1-i)^2}}} .......first simplify

{{{1/(1^2-2i+i^2)}}}

{{{1/(1-2i-1)}}}

{{{-1/2i}}}........rationalize

{{{-1i/(2i*i)}}}

{{{-i/(2i^2)}}}

{{{-i/(-2)}}}

{{{i/2}}}

so {{{z=i/2}}}

Polar form 

{{{r*arg(z)=r(cos(theta)+i*sin(theta))}}}

where {{{r= sqrt(x ^2+y^2)}}} and where {{{z=a+bi}}}


=> {{{a=0 }}}and {{{b=1/2}}}


{{{ r= sqrt(0 ^2+(1/2)^2)}}}

{{{r= 1/2}}}


To find argument  we use one of the following formulas:

 {{{theta=tan^-1(b/a)}}} if {{{a>0}}}

{{{theta=tan^-1(b/a) +180}}}° if {{{ a<0}}}

{{{theta=90}}}° if {{{a=0}}} and {{{b>0}}}

 in your case {{{a=0}}} and {{{ b=1/2}}} , so {{{theta=90}}}°

and polar form is:

{{{(1/2) (cos(90)+i*sin(90) )}}}