We have,
\(|w|=1\ \Rightarrow \frac { |1-iz| }{ |z-i| } =1\)
\(\Rightarrow |1-iz|=|z-i|\)
\(\Rightarrow |1+y-ix|=|x+i(y-1)|\)

Let z = x+ iy, then
\(|x+iy+1|=x+iy+2(1+i)\)
\( \Rightarrow \sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } =x+2+i(y+2)\)
\(z=\frac { 1 }{ 2 } -2i\)

Here z = -3 = r (cos\(\theta \) + i sin\(\theta \))
⇒ cos\(\theta \) =-3 and r sin\(\theta \)= 0 ...(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \) + i sin²\(\theta \))= 9+0
⇒ r²= 9 ⇒r =3
∴ 3 cos\(\theta \)= - 3 and 3 sin\(\theta \)= 0
⇒ cos\(\theta \)= -1 and sin\(\theta \)=0
Since sin\(\theta \) is positive and cos\(\theta \) is negative
∴ \(\theta \) lies in second quadran t
∴   \(\theta \) = (\(\pi \) - 0) = \(\pi \)
Hence polar form of z is 3(cos \(\pi \)+ i sin \(\pi \)).

Here z= -1+i=r(cos\(\theta \)+i sin\(\theta \))
⇒ rcos \(\theta \)= -1 and r sin\(\theta \)=1   ..(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \)) = 1+1
∴  \(\sqrt { 2 } \)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 
∴  \(\theta =\left( \pi -\frac { \pi }{ 4 } \right) =\frac { 3\pi }{ 4 } \)
Hence polar form of z is 
\(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } +i\quad sin\frac { 3\pi }{ 4 } \right) \)

We have, a + ib = \(\frac { ({ x }^{ 2 }+1) }{ 2{ x }^{ 2 }+1 } \) ....(i)
Take modulus both sides of E_q . (i) and then solve it.

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, \(y=-\sqrt 3\)
\(\therefore\sqrt {-2+2\sqrt{3}i}=\pm(1+\sqrt 3i)\).

Let \(x+yi=\sqrt {-8i}\)
Squaring both sides, we get
x²-y²+2xyi=-8i
Comparing the real and imaginary parts
x²-y²=0......(i)
2xy=-8 \(\Rightarrow\) xy=-4
Now from the identity we have
(x²+y²)²=(x²-y²)²+4x²y²
=(0)²+4(-4)²
=64
\(\therefore\)x²+y²=8........(ii)[Neglecting (-) sign as x²+y²>0]
Solving (i) and (ii), we get
x²=4 and y²=4
\(\therefore\)x=士2 and y=士2
Since the sign of xy is (-)
\(\therefore\) if x=2,y=-2
and if x=-2,y=2
\(\therefore\sqrt {-8i}=\pm(2-2i).\)

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= \(\left( \frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } \right) ^{ 2 }+\left( \frac { 2x }{ { 2x }^{ 2 }+1 } \right) ^{ 2 }\)
= \(\frac { { (x }^{ 2 }-1)^{ 2 } }{ (2{ x }^{ 2 }+1)^{ 2 } } +\frac { (2x)^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { (x }^{ 2 }-1)^{ 2 }+(2x)^{ 2 } }{ (2{ x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { x }^{ 4 }+1-{ 2x }^{ 2 }+{ 4x }^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { x }^{ 4 }+1+{ 2x }^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } =\frac { ({ x }^{ 2 }+1)^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)

Let \(x+yi=\sqrt {-5+12i}\)
Squaring both sides, we get
x² - y² + 2xyi = -5 + 12i
Equating the real and imaginary parts
x²-y²=-5......(i)
2xy = 12 \(\Rightarrow\) xy = 6
Now from the identity, we have
(x² + y²)² = (x² - y²)² + 4x²y²
=(-5)² + 4(6)² = 25 + 144 = 169
\(\therefore\) x² + y² = 13....(ii)  [Neglecting (-) sign as x²+y²>0]
Solving (i) and (ii) we get
x² = 4 and y² = 9
\(\therefore\) x = 土2 and y = 土3
Since the sign of xy is (+),
\(\therefore\) if x = 2,y = 3
and if x = -2,y = -3
\(\therefore\sqrt {-5+12i}=\pm(2+3i)\)

Here \(\frac{1-ix}{1+ix}=a-ib\) By componendo and dividendo,
we have \(\frac{1-ix+1+ix}{1-ix-1-ix}=\frac{a-ib+1}{a-ib-1}\)
\(\Rightarrow \frac{2}{-2ix}=\frac{1+a-ib}{-(1-a+ib)}\)
\(\Rightarrow \frac{1}{ix}=\frac{1+a-ib}{1-a+ib}\)
\(\Rightarrow ix=\frac{1-a+ib}{1+a-ib}\times\frac{1+a+ib}{1+a+ib}\)
\(\Rightarrow ix=\frac{1-a^2-b^2+2ib}{(1+a)^2-i^2b^2}\)
\(\Rightarrow ix=\frac{1-a^2-b^2+2ib}{(1+a)^2+b^2}\)
\(=\frac{1-a^2-b^2}{(1+a)^2+b^2}+\frac{2b}{(1+a)^2+b^2}i\)
If a² + b² = 1 then
\(x=\frac{2b}{(1+a)^2+b^2}\) which is real.

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
\((x+iy)(x-iy)=\frac{(a+i)^2}{2a-i}\times\frac{(a-i)^2}{2a+i}\)
\(\Rightarrow =x^2-i^2y^2=\frac{(a+i)^2(a-i)^2}{4a^2-i^2}\)
\(\Rightarrow =x^2+y^2=\frac{(a^2-i^2)^2}{4a^2+1}\ [\because i^2=-1]\)
\(=\frac{(a^2+1)^2}{4a^2+1}\)​​​​​​​