Here a+ib =\(\frac { x+i }{ x-i } =\frac { x+i }{ x-i } \times \frac { x+i }{ x+i } \)
= \(\frac { (x+i)^{ 2 } }{ x^{ 2 }-i^{ 2 } } =\frac { { x }^{ 2 }+2xi+{ i }^{ 2 } }{ { x }^{ 2 }+1 } \)
= \(\frac { x^{ 2 }-1 }{ x^{ 2 }+1 } +\frac { 2x }{ { x }^{ 2 }+1 } i\)
Comparing real and imaginary parts on both sides, we have
a=\(\frac { x^{ 2 }-1 }{ x^{ 2 }+1 } \)and b=\(\frac { 2x }{ { x }^{ 2 }+1 } \)
Now a²+b²= \(\left( \frac { x^{ 2 }-1 }{ x^{ 2 }+1 } \right) ^{ 2 }+\left( \frac { 2x }{ { x }^{ 2 }+1 } \right) ^{ 2 }\)
= \(\frac { (x^{ 2 }-1)^{ 2 }+{ 4x }^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { ({ x }^{ 2 }+1)^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 } } =1\)
Also \(\frac { b }{ a } =\frac { \frac { 2x }{ { x }^{ 2 }+1 } }{ \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } } =\frac { 2x }{ { x }^{ 2 }-1 } \)

Now \(\left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| ^{ 2 }\)
= \(\left| \frac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right| \left| \frac { \overline { \beta -\alpha } }{ \overline { 1-\overline { \alpha } \beta } } \right| \)              \(\left[ \because |z|^{ 2 }=z\overline { z } \right] \)  
= \(\left| \frac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right| \left| \frac { \overline { \beta } -\overline { \alpha } }{ \overline { 1-\alpha \overline { \beta } } } \right| \)
=  \(\frac { \beta \overline { \beta } -\beta \overline { \alpha } -\alpha \overline { \beta } +\alpha \overline { \alpha } }{ 1-\overline { \alpha } \beta -\alpha \overline { \beta } +\alpha \overline { \alpha } \beta \overline { \beta } } \)
= \(\frac { |\beta |^{ 2 }-\overline { \alpha } \beta -\alpha \overline { \beta } +|\alpha |^{ 2 } }{ 1-\overline { \alpha } \beta -\alpha \overline { \beta } +|\alpha |^{ 2 }|\beta |^{ 2 } } \)
= \(\frac { 1-\overline { \alpha } \beta -\alpha \overline { \beta } +|\alpha |^{ 2 } }{ 1-\overline { \alpha } \beta -\overline { \alpha } \beta +|\alpha |^{ 2 } } =1\)
∴\(\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|\) =1

Let \(x+yi=\sqrt {-2+2\sqrt {3}i}\)
Squaring both sides, we get
x²-y²+2xyi=\(-2+2\sqrt {3}i\)
Comparing the real and imaginary parts
x²-y²=-2.....(i)
2xy=\(2\sqrt 3\)  \(\Rightarrow\) xy = \(\sqrt 3\)
Now, from the identity, we know
(x²+y²)²=(x²-y²)²+4x²y²
=(-2)²+4(\(\sqrt 3\))2=4+12=16
\(\therefore x^2+y^2=4\)........(ii)[neglecting (-) sign as x²-y²>0]
Solving (i) and (ii), we get
x²=1 and y²=3
\(\therefore x=\pm1\ and y=\pm\sqrt 3\)
Since the sign of xy is (+)
\(\therefore\) if x=1,\(y=\sqrt 3\)
and if x=-1, \(y=-\sqrt 3\)
\(\therefore\sqrt {-2+2\sqrt{3}i}=\pm(1+\sqrt 3i)\).

(i) Here z₁= 2-i and z₂=-2+i
∴ \(\bar { { Z }_{ 1 } } \) =2+i
z₁z₂ =(2-i)(-2+i)=-4+2i+2i-i²
=(-4+1)+4i=-3+4i
∴  = \(\frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { Z }_{ 1 } } } =\frac { -3+4i }{ 2+i } \div \times \frac { 2-i }{ 2-i } \)
= \(\frac { -6+3i+8i-4{ i }^{ 2 } }{ 4-{ i }^{ 2 } } \)
= \(\frac { (-6+4)+11i }{ 4+1 } =\frac { -2+11i }{ 5 } \)
= \(\frac { -2 }{ 5 } +\frac { 11 }{ 5 } i\)
∴ Re\(\left( \frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } \right) \) =\(\frac { -2 }{ 5 } \)
(ii) \(\frac { 1 }{ { z }_{ 1 }\bar { { z }_{ 1 } } } =\frac { 1 }{ (2-i)(2+i) } =\frac { 1 }{ 4-{ i }^{ 2 } } =\frac { 1 }{ 5 } \)
On comparing imaginary parts , we obtain
∴ \(Im\left( \frac { 1 }{ { z }_{ 1 }\bar { { z }_{ 1 } } } \right) =0\)

Here a+ib =\(\frac { ({ x }^{ 2 }+i)^{ 2 } }{ { 2x }^{ 2 }+1 } =\frac { { x }^{ 2 }+i+2ix }{ { 2x }^{ 2 }+1 } \)
=  \(\frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } +i\frac { 2x }{ { 2x }^{ 2 }+1 } \)
Comparing both sides, we have
a= \(\frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } \)and b= \(\frac { 2x }{ { 2x }^{ 2 }+1 } \)
∴ a²+b²= \(\left( \frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } \right) ^{ 2 }+\left( \frac { 2x }{ { 2x }^{ 2 }+1 } \right) ^{ 2 }\)
= \(\frac { { (x }^{ 2 }-1)^{ 2 } }{ (2{ x }^{ 2 }+1)^{ 2 } } +\frac { (2x)^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { (x }^{ 2 }-1)^{ 2 }+(2x)^{ 2 } }{ (2{ x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { x }^{ 4 }+1-{ 2x }^{ 2 }+{ 4x }^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { x }^{ 4 }+1+{ 2x }^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } =\frac { ({ x }^{ 2 }+1)^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)

\(\frac { 1+3i }{ 1-2i } \times \frac { 1+2i }{ 1+2i } =\frac { 1+2i+3i+6{ i }^{ 2 } }{ 1-4{ i }^{ 2 } } \)
=\(\frac { -5+5i }{ 5 } \)=-1+i
Let z = -1+i=r(cos\(\theta \)+i sin\(\theta \))
⇒ r cos\(\theta \)=-1 and r sin\(\theta \) =1   ...(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \))=1+1
⇒  r²=2 ⇒ r= \(\sqrt { 2 } \)
∴ \(\sqrt { 2 } \)cos\(\theta \)=-1 and \(\sqrt { 2 } \)sin\(\theta \)=1
⇒ cos\(\theta \) =\(\frac { -1 }{ \sqrt { 2 } } \) and sin \(\theta \)=\(\frac { -1 }{ \sqrt { 2 } } \)
Since sin\(\theta \) is positive and cos\(\theta \) is negative
∴ \(\theta \) lies in second quadrant
∴    \(\theta \)=\(\pi -\frac { \pi }{ 4 } =\frac { 3\pi }{ 4 } \)
Hence polar form of z is \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } +isin\frac { 3\pi }{ 4 } \right) \)

Here \(x+iy=\frac{(a+i)^2}{2a-i}\).......(i)
Taking conjugate on both sides, we have
\(\overline {x+iy}=\frac{\overline {(a+i)}^2}{\overline {(2a-i)}}\)
\(\Rightarrow =x-iy=\frac{(a-i)^2}{2a+i}\).....(ii)
Multiplying (i) and (ii), we have
\((x+iy)(x-iy)=\frac{(a+i)^2}{2a-i}\times\frac{(a-i)^2}{2a+i}\)
\(\Rightarrow =x^2-i^2y^2=\frac{(a+i)^2(a-i)^2}{4a^2-i^2}\)
\(\Rightarrow =x^2+y^2=\frac{(a^2-i^2)^2}{4a^2+1}\ [\because i^2=-1]\)
\(=\frac{(a^2+1)^2}{4a^2+1}\)​​​​​​​

Here z=\(\sqrt { 3 } \)+1 = r (cos\(\theta \)+i sin\(\theta \))
⇒ r cos\(\theta \) =\(\sqrt { 3 } \) and r sin\(\theta \) =1   ..(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \))=3+1
⇒ r²= 4⇒ r =2
∴  2 cos \(\theta \)=\(\sqrt { 3 } \)and 2 sin \(\theta \)=1
∴  cos\(\theta \)=\(\frac { \sqrt { 3 } }{ 2 } \)and sin\(\theta \) =\(\frac { 1 }{ 2 } \)
Since sin\(\theta \) and cos\(\theta \) are both positive
∴  \(\theta \) lies in first quadrant
∴   \(\theta \)  = \(\frac { \pi }{ 6 } \)
Hence polar form of z is 2\(\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)

Here z= -\(\sqrt { 3 } \)+i=r(cos\(\theta \)+i sin\(\theta \))
⇒ r cos\(\theta \)= and r sin\(\theta \)=1 ...(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \))=3+1
⇒ r²=4 ⇒ r =2
∴ 2cos\(\theta \) =- \(\sqrt { 3 } \)and 2sin\(\theta \) =1
⇒ cos\(\theta \) =\(\frac { -\sqrt { 3 } }{ 2 } \) and sin \(\theta \)=\(\frac { 1 }{ 2 } \)
Since sin\(\theta \) is positive and cos\(\theta \) is negative
∴ \(\theta \) lies in second quadrant
∴ \(\theta =\left( \pi -\frac { \pi }{ 6 } \right) =\frac { 5\pi }{ 6 } \)  
∴ \(\left| z \right| \) =2 and arg (z)=\(\frac { 5\pi }{ 6 } \)

Here z = -1-\(\sqrt { 3 } \) = r(cos\(\theta \) + i sin\(\theta \))
⇒ r cos =-1 and r sin =-\(\sqrt { 3 } \)    ..(i)
Squaring both sides of (i) and adding
r² (cos²+sin²) = 1+3
⇒ r² = 4 ⇒r =2
∴ 2 cos \(\theta \) = -1 and 2sin\(\theta \) = \(\sqrt { 3 } \)
⇒ cos \(\theta \)=\(\frac { -1 }{ 2 } \)and sin \(\theta \)=\(\frac { -\sqrt { 3 } }{ 2 } \)
Since both sin\(\theta \) and cos\(\theta \) are negative
∴ \(\theta \) lies in third quadrant.
∴ \(\theta \)= \(\left( -\pi +\frac { \pi }{ 3 } \right) =\frac { -2\pi }{ 3 } \)
∴  \(|z|=2\quad and\quad arg(z)=\frac { -2\pi }{ 3 } \)