\(\frac { { \left( 1-i \right) }^{ 3 } }{ 1-{ i }^{ 3 } } =\frac { { 1 }^{ 3 }-{ i }^{ 3 }-3i+3{ i }^{ 2 } }{ 1+i } =\frac { 1+i-3i-3 }{ 1+i } =\frac { -2-2i }{ 1+i } \)
Ans.-2

\(z=\frac { 1+icos\theta }{ 1-2icos\theta } \times \frac { 1+2icos\theta }{ 1+2icos\theta } =\frac { 1-2{ cos }^{ 2 }\theta +3icos\theta }{ 1+4{ cos }^{ 2 }\theta } \)
\(For\quad Re(z),put\quad 3cos\theta =0\Rightarrow cos\theta =0\)
Ans. \(2n\pi \pm \frac { \pi }{ 2 } \)

We have, \({ \left( x+iy \right) }^{ 1/3 }=a+ib\) 
\(\Rightarrow x+iy={ \left( a+ib \right) }^{ 3 }\quad \left[ cubing\quad both\quad sides \right] \)
\(\Rightarrow x+iy={ a }^{ 3 }+{ i }^{ 3 }{ b }^{ 3 }+3{ a }^{ 2 }bi+3a{ b }^{ 2 }i^{ 2 }\)
\(\left[ \because { \left( { z }_{ 1 }+{ z }_{ 2 } \right) }^{ 3 }={ z }_{ 1 }^{ 3 }+{ z }_{ 2 }^{ 3 }+3{ z }_{ 1 }^{ 2 }{ z }_{ 2 }+{ 3z }_{ 1 }{ z }_{ 2 }^{ 2 } \right] \)
\(\Rightarrow x+iy={ a }^{ 3 }+i{ b }^{ 3 }+i3{ a }^{ 2 }b-3a{ b }^{ 2 }\)
\(\left[ \because { i }^{ 3 }=-i\ and\ { i }^{ 2 }=-1 \right]\)
\(\Rightarrow x+iy={ a }^{ 3 }-3a{ b }^{ 2 }+i\left( 3{ a }^{ 2 }b-{ b }^{ 3 } \right) \)
On equating real and imaginary parts from both sides, we get
\(x={ a }^{ 3 }-3a{ b }^{ 2 }\quad and\quad y=\left( 3{ a }^{ 2 }b-{ b }^{ 3 } \right) \)  
\(\Rightarrow \quad \frac { x }{ a } ={ a }^{ 2 }-{ 3b }^{ 2 }\quad and\quad \frac { y }{ b } ={ 3a }^{ 2 }-{ b }^{ 2 }\) 
Now, \(\frac { x }{ a } -\frac { y }{ b } ={ a }^{ 2 }-{ 3b }^{ 2 }-{ 3a }^{ 2 }+{ b }^{ 2 }\) 
\(=-2{ a }^{ 2 }-2b^{ 2 }=-2\left( { a }^{ 2 }+{ b }^{ 2 } \right) \)

We have,
\({ \left( 1+i \right) }^{ 6 }={ \left( { \left( 1+i \right) }^{ 2 } \right) }^{ 3 }\)
\(={ \left( 1-{ i }^{ 2 }+2i \right) }^{ 3 }\)
\(\left[ \because { \left( { z }_{ 1 }+{ z }_{ 2 } \right) }^{ 2 }={ z }_{ 1 }^{ 2 }+{ z }_{ 2 }^{ 2 }+{ 2z }_{ 1 }{ z }_{ 2 } \right] \)
\(={ \left( 1-1+2i \right) }^{ 3 }\ \left[ \because { i }^{ 3 }=-1 \right] \)
\(\Rightarrow { \left( 1+i \right) }^{ 6 }={ \left( 2i \right) }^{ 3 }={ 8 }i^{ 3 }=8i \left[ \because { i }^{ 3 }=-1 \right] \left( i \right) \)
\(and\quad { \left( 1-i \right) }^{ 3 }={ i }^{ 3 }-{ i }^{ 3 }-3{ \left( 1 \right) }^{ 2 }i+3\left( 1 \right) { \left( i \right) }^{ 2 }\)
\(\left[ \because { \left( { z }_{ 1 }-{ z }_{ 2 } \right) }^{ 2 }={ z }_{ 1 }^{ 3 }-{ 3z }_{ 1 }^{ 2 }{ z }_{ 2 }+3{ z }_{ 1 }{ z }_{ 2 }^{ 2 }-{ z }_{ 2 }^{ 3 } \right] \)
\(=1-\left( -i \right) -3i-3\ \left[ \because { i }^{ 3 }=-1\ { i }^{ 2 }=-1 \right]\) 
\(\Rightarrow(1-i)^{3}=-2-2 i\)
On adding eqs. (i) and (ii), we get
(1 + i)⁶ + (1-i)³ 
= - 8i - 2 - 2i
= -2 - 10i.

\(z=\frac { 1-i }{ 1+i } x \frac { 1-i }{ 1-i } =\frac { 1-1-2i }{ 1+1 } =-i\)   = i

We have, (1+ i)z = \((1-i) \bar{z}\)
\(\frac{z}{\bar{z}}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}=\frac{1-1-2 i}{1+1}\)