Answer
$$\frac{1}{4} + \frac{1}{4}i$$
Work Step by Step
$$\eqalign{
& \frac{1}{{2 - 2i}} \cr
& {\text{Convert the numerator and the denominator to trigonometric form}} \cr
& 1 = \left( {\cos {0^ \circ } + i\sin {0^ \circ }} \right) \cr
& 1 + i = 2\sqrt 2 \left( {\cos {{315}^ \circ } + i\sin {{315}^ \circ }} \right) \cr
& {\text{Then,}} \cr
& \frac{{ - i}}{{1 + i}} = \frac{{\cos {0^ \circ } + i\sin {0^ \circ }}}{{2\sqrt 2 \left( {\cos {{315}^ \circ } + i\sin {{315}^ \circ }} \right)}} \cr
& {\text{Using the Quotient Theorem}} \cr
& = \frac{1}{{2\sqrt 2 }}\left[ {\cos \left( {{0^ \circ } - {{315}^ \circ }} \right) + i\sin \left( {{0^ \circ } - {{315}^ \circ }} \right)} \right] \cr
& {\text{Simplify}} \cr
& = \frac{1}{{2\sqrt 2 }}\left[ {\cos \left( { - {{315}^ \circ }} \right) + i\sin \left( { - {{315}^ \circ }} \right)} \right] \cr
& = \frac{1}{{2\sqrt 2 }}\left[ {\cos \left( {{{315}^ \circ }} \right) - i\sin \left( {{{315}^ \circ }} \right)} \right] \cr
& {\text{Write in Rectangular form}} \cr
& = \frac{1}{{2\sqrt 2 }}\left( {\frac{{\sqrt 2 }}{2} - \left( { - \frac{{\sqrt 2 }}{2}} \right)i} \right) \cr
& = \frac{1}{4} + \frac{1}{4}i \cr} $$
