Consider, \(\frac { { i }^{ 4x+1 }-{ i }^{ 4x-1 } }{ 2 } =\frac { { i }^{ 4x }.i-{ i }^{ 4x }.{ i }^{ -1 } }{ 2 } =\frac { i-\frac { 1 }{ i } }{ 2 } \quad [\because { i }^{ 4x }=1]\)
\(=\frac { { i }^{ 2 }-1 }{ 2i } =\frac { -2 }{ 2i } \quad [\because { i }^{ 2=-1 }]\)
\(=\frac { -1 }{ i } =\frac { -i }{ { i }^{ 2 } } =\frac { -i }{ -1 } =i\quad [\because { i }^{ 2 }=-1]\)
  

\(\left( 3+\sqrt { -5 } \right) \left( 3-\sqrt { -5 } \right) = \left( 3+\sqrt { 5i } \right) \left( 3-\sqrt { 5i } \right) \)
\(={ \left( 3 \right) }^{ 2 }-{ \left( \sqrt { 5i } \right) }^{ 2 }\quad \left[ \therefore \left( { z }_{ 1 }-{ z }_{ 2 } \right) \left( { z }_{ 1 }{ +z }_{ 2 } \right) ={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } \right]\)
\(=9-{ 5i }^{ 2 }=9+5=14\)

\({ \left( \frac { 1 }{ 3 } +3i \right) }^{ 3 }=\quad { \left( \frac { 1 }{ 3 } \right) }^{ 3 }+3{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }\left( 3i \right) +3{ \left( \frac { 1 }{ 3 } \right) }{ \left( 3i \right) }^{ 2 }+{ \left( 3i \right) }^{ 3 }\)
\(\left[ \therefore \ { \left( { z }_{ 1 }+{ z }_{ 2 } \right) }^{ 3 }={ z }_{ 1 }^{ 3 }+3{ z }_{ 1 }^{ 2 }{ z }_{ 2 }+{ z }_{ 2 }^{ 3 } \right] \)
\(=\frac { 1 }{ 27 } +3\left( \frac { 1 }{ 9 } \right) \left( 3i \right) +3\left( \frac { 1 }{ 3 } \right) \left( { 9i }^{ 2 } \right) +{ 27i }^{ 3 }\)
\(=\frac { 1 }{ 27 } +i+9\left( -1 \right) +27\left( -i \right) \quad \left[ \therefore { i }^{ 2 }=-1and\quad { i }^{ 3 }=-i \right] \)
\(=\frac { 1 }{ 27 } -9-26i=\frac { 1-243 }{ 27 } -26i=\frac { -242 }{ 27 } -26i\)

\(\left( -i \right) \left( 3i \right) { \left( -\frac { 1 }{ 6 } i \right) }^{ 3 }=\quad \left( -3i \right) \left( -\frac { 1 }{ 216 } { i }^{ 3 } \right) \)
\(=\left( -3\times \left( -1 \right) \right) \left( -\frac { 1 }{ 216 } \left( -i \right) \right) \quad \left[ \because { i }^{ 2 }=-1\quad and\quad { i }^{ 3 }=-i \right] \)
\(=3\times \frac { 1 }{ 216 } \times i\)
\(=\frac { i }{ 72 } =0+\frac { 1 }{ 72 } i\) 
Which is in form of a + ib.

We have, (x² + 2xi) - (3x² + iy ) = (3 - 5i) + (1 + 2iy)
\(\Rightarrow \) (x⁴ - 3x²) + (2x - y)i = 4 + (-5 + 2y)i
On equating real and imaginary parts both sides, we get
x⁴ - 3x² = 4 ...(i)
and  2x - y = -5 + 2y
 \(\Rightarrow \)  2x - 3y = -5......(ii)
On solving Eq. (i), we get
x⁴ - 3x² = 4 \(\Rightarrow \) x⁴ - 3x² - 4 = 0
\(\Rightarrow \)  x⁴ - 4x² + x² - 4 = 0
\(\Rightarrow \) (x² - 4)(x² + 1) = 0 \(\Rightarrow \) x² - 4 = 0
[\(\therefore \) x² + 1 \(\neq \) 0, for any real value of x]
\(\therefore \)  x =\(\pm \)2
On putting x =\(\pm \)2 in Eq. (ii), we get
y = 3, when x =2 and y = \(\frac { 1 }{ 3 } \), when x = -2
Thus, x = -2, y = \(\frac { 1 }{ 3 } \) or x = 2, y = 3.

\(\frac{9}{16}\pm\frac{\sqrt {15}}{16}i\)

\(\frac{3}{5}\pm\frac{1}{5}i\)

\(|z|=\sqrt 2,arg(z)=\frac{3\pi}{4};\sqrt 2(cos\frac{3\pi}{4}+isin\frac{3\pi}{4})\)

Let  z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂​
Then, ​​​​​​z₁z₂ = (x₁x₂ - y₁y₂) + i(x₁x₂ + y₁y₂)
\(\therefore \)  Re(z₁z₂) = x₁x₂ - y₁y₂
=Re(z₁) Re(z₂) - Im(z₁) Im(z₂)
Hence Proved.

We have, \(a+ib=\frac { x+i }{ x-{ i } } =\frac { x+i }{ x-{ i } } \times \frac { x+i }{ x+i } \)
[by rationalising the denominator]
\(=\frac { { x }^{ 2 }+2xi+{ i }^{ 2 } }{ { x }^{ 2 }-{ i }^{ 2 } } =\frac { { x }^{ 2 }-1+2xi }{ { x }^{ 2 }+1 } \quad \left[ \because \ { i }^{ 2 }=-1 \right] \)
\(\Rightarrow a+ib=\frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } +\frac { 2x }{ { x }^{ 2 }+1 } i\)
On comparing real and imaginary parts both sides, we get
\(a=\frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }\quad and\quad b=\frac { 2x }{ { x }^{ 2 }+1 }\) ..(i)
Now, 
\({ a }^{ 2 }+b^{ 2 }={ \left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right) }^{ 2 }+{ \left( \frac { 2x }{ { x }^{ 2 }+1 } \right) }^{ 2 } \quad [from\ Eq.(i)]\)
\(=\frac { { \left( { x }^{ 2 }-1 \right) }^{ 2 }+4{ x }^{ 2 } }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } } =\frac { { x }^{ 4 }+1-2{ x }^{ 2 }+4{ x }^{ 2 } }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } } \)
\(=\frac { { x }^{ 4 }+1+2{ x }^{ 2 } }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } } =\frac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } } =1\)
\(Also,\ \frac { b }{ a } =\frac { \frac { 2x }{ { x }^{ 2 }+1 } }{ \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } } =\frac { 2x }{ { x }^{ 2 }-1 } \ [from\ Eq.(i)]\)
Hence proved.