(-3i) (i)\(\left( -\frac { 1 }{ 4 } i \right) ^{ 3 }\)
= -3i²\(\times -\frac { 1 }{ 64 } { i }^{ 3 }=\frac { 3 }{ 64 } { i }^{ 5 }\)
= \(\frac { 3 }{ 64 } \)(i²)².i=\(\frac { 3 }{ 64 } \)i= 0+ \(\frac { 3 }{ 64 } \)i

\(\pm(1-i)\)

\(\frac{1}{2197}(-46+9i)\)

\((4\sqrt 2-6)+2\sqrt 3(2+\sqrt 2)i\)

sec∝(cos∝+i sin∝)

\(8,\frac { 2\pi }{ 3 } \)

-2

\(\left[ \frac { 1-cos\theta }{ 2-2cos\theta +3sin^{ 2 }\theta } \right] +i\left[ \frac { -2sin\theta }{ 2-2cos\theta +3sin^{ 2 }\theta } \right] \)

\(\frac { 63 }{ 25 } -\frac { 16 }{ 25 } i\)

\(\frac { 40 }{ 41 } -\frac { 9 }{ 41 } i\)

 \(\left( \frac { 1 }{ 5 } +\frac { 2 }{ 5 } i \right) -\left( 4+\frac { 5 }{ 2 } i \right) \)
= \(\frac { 1 }{ 5 } +\frac { 2 }{ 5 } i-4-\frac { 5 }{ 2 } i\)
= \(\left( \frac { 1 }{ 5 } -4 \right) +\left( \frac { 2 }{ 5 } -\frac { 5 }{ 2 } \right) i\)
= \(\frac { -19 }{ 5 } -\frac { 21 }{ 10 } i\)

3(7+i7)+i(7+i7) = 21+21i+7i+7i²
= 21+28i-7 =14+28i

Let z = (3 - 2i) (3 + 2i) (1 + i)
z = (9 = 6i - 6i - 4i²) (1 + i)
= ( 9 + 4) (1 + i) = 13 + 13i
\(\overline { z } =13-13i\ and\ \left| z \right| =13\sqrt { 2 } \) 

z = (6 + 5i )² = 36 - 25 + 60i = 11 + 60i 
= 11 - 60i

\(z=\frac { 2+3i }{ 3+2i } x \frac { 3-2i }{ 3-2i } =\frac { 12+5i }{ 13 } \)
\(\overline { z } =\frac { 12 }{ 13 } -\frac { 5 }{ 13 } i\quad and\quad \left| z \right| =1\)