We know that, when two complex numbers are equal, then their conjugates are also equal.
(a + ib)² = x+iy .........(1)
(a - ib)² = x-iy ..........(2)
Multiplying, (a + ib)² (a - ib)² = (x + iy) (x - iy)
(i) \( {[(a+i b)(a-i b)]^{2} } =x^{2}+y^{2} \)
\(\left(a^{2}+b^{2}\right)^{2} =x^{2}+y^{2} \)
(ii) \( (\cos \theta-i \sin \theta)^{2}=x-i y \)
\(\Rightarrow \ (\cos \theta+i \sin \theta)^{2}=x+i y\)
Multiplying
\( (\cos \theta-i \sin \theta)^{2}(\cos \theta+i \sin \theta)^{2} =(x-i y)(x+i y) \)
\({[(\cos \theta-i \sin \theta)(\cos \theta+i \sin \theta)]^{2} } =x^{2}+y^{2} \)
\({\left[\cos ^{2} \theta+\sin ^{2} \theta\right]^{2} } =x^{2}+y^{2} \)
\(x^{2}+y^{2} =1\)
Hence proved.

\(p+i q=\frac{(a+i)^{2}}{2 a-i}\) ..............(1)
Replacing 'i' by '- i' we get
\(p-i q=\frac{(a-i)^{2}}{2 a+i}\) ...............(2)
Multiplying (1) and (2)
\( (p+i q)(p-i q) =\frac{(a+i)^{2}}{2 a-i} \times \frac{(a-i)^{2}}{2 a+i} \)
\(p^{2}-(i q)^{2} =\frac{[(a+i)(a-i)]^{2}}{(2 a)^{2}-i^{2}} \)
\(p^{2}+q^{2} =\frac{\left(a^{2}+1\right)^{2}}{4 a^{2}+1} \)
Hence proved.

z = 3+5i, \(\bar z\) = 3-5i
\( z^{3} =(3+5 i)^{3} \)
\(=(3)^{3}+(5 i)^{3}+3(3)(5 i)(3+5 i)\)
= 27 - 125i + 45i (3 + 5i)
= 27 -125i + 135i -225
= -198 +10i
Now \(z^{3}+\bar{z}+198\) = - 198 + 10i + 3 -5i + 198
= 3 + 5i = z

\( x+\mathrm{iy} =\frac{a+i}{a-i} \times \frac{a+i}{a+i} \)
\(=\frac{(a+i)^{2}}{a^{2}+1}=\frac{a^{2}-1+2 a i}{a^{2}+1} \)
Comparing real and imaginary parts
\( x=\frac{a^{2}-1}{a^{2}+1}\ \text {and } y=\frac{2 a}{a^{2}+1} \)
Now, \( ay -1=a\left(\frac{2 a}{a^{2}+1}\right)-1 \)
\(=\frac{2 a^{2}}{a^{2}+1}-1=\frac{2 a^{2}-a^{2}-1}{a^{2}+1} \)
\(=\frac{a^{2}-1}{a^{2}+1}=x\)
ay - 1 = x
Hence proved.

Let z = 1 - i 
\(\therefore\) Additive inverse of z is - z
Additive inverse of 1 - i is -(1 -i) = - 1 + i
Multiplicative inverse of z is z^-1 
we know z z^-1 = 1 
Let z^-1 be a + ib
\(\therefore\) (1-i)(a+ib) = 1 
a + ib - ia + b = 1 
(a + b) + i(b - a) = 1+i0
Equating real and imaginary parts
a + b = 1, b - a = 0
Solving these equations
\( a=\frac{1}{2}, b=\frac{1}{2} \)
\(\therefore\) These inverse of \(1-i=\frac{1}{2}+\frac{1}{2} i\)

\( (x+i y)^{\frac{1}{3}} =a+i b \)
\(x+i y =(a+i b)^{3} \)
\(x+i y =a^{3}+3 a^{2}(i b)+3 a(i b)^{2}+(i b)^{3} \)
\(x+i y =a\left(a^{2}-3 b^{2}\right)+i b\left(3 a^{2}-b^{2}\right)\)
Equating real and imaginary parts
\( x=a\left(a^{2}-3 b^{2}\right), \quad y=b\left(3 a^{2}-b^{2}\right) \)
\( \frac{x}{a}=a^{2}-3 b^{2}, \quad \frac{y}{b}=3 a^{2}-b^{2} \)
\( \therefore \frac{x}{a}+\frac{y}{b}=a^{2}-3 b^{2}+3 a^{2}-b^{2} \)
\( =4 a^{2}-4 b^{2}=4\left(a^{2}-b^{2}\right) \)
Hence proved.

\(\left(\frac{3}{\sqrt{5}} x-5\right)+i 2 \sqrt{5} y=\sqrt{2}\)
Equating real and imaginary parts
\( \frac{3}{\sqrt{5}} x-5 =\sqrt{2}, \ 2 \sqrt{5} y=0 \)
\(\frac{3}{\sqrt{5}} x =\sqrt{2}+5 \ \ \ y=0 \)
\(x =\frac{\sqrt{5}(\sqrt{2}+5)}{3} \)
\(x =\frac{\sqrt{5}(\sqrt{2}+5)}{3} \ \text { and } y=0 \)

x + 4iy = ix + y + 3 
26 + 4iy = (y+3) + ix
Equating real and imaginary parts x = y + 3 and x - 4y
Solving the equations
4y = y + 3
3y = 3
y = 1
x = 4(1) = 4
x = 4, y = 1

\( {\left[i^{17}-\left(\frac{1}{i}\right)^{3}\right]^{2} } =\left[i^{16} i-\left(\frac{1}{i^{32} i^{2}}\right)\right]^{2} \)
\( =\left[\left(i^{4}\right)^{4} i-\frac{1}{\left(i^{4}\right)^{8} i^{2}}\right]^{2} \)
\( =[\mathrm{i}+1]^{2}=\mathrm{i}^{2}+2 \mathrm{i}+1 \)
\( =-1+2 \mathrm{i}+1=2 \mathrm{i}\)

L.H.S. = \( (1+i)^{4}\left(1+\frac{1}{i}\right)^{4}=(1+i)^{4}\left(1+\frac{1}{i}\right)^{4} \)
\( =(1+i)^{4}\left(1+\frac{i}{i^{2}}\right)^{4}\left[\because i^{2}=-1\right] \)
\( =(1+i)^{4}(1-i)^{4} \)
\( =[(1+\mathrm{i})(1-\mathrm{i})]^{4} \)
\( =\left(1-\mathrm{i}^{2}\right)^{4}=(1+1)^{4}=2^{4}=16\)

\( i^{n+100}+i^{n+50}+i^{n+48}+i^{n+46}=i^{n}\left[i^{100}+i^{50}+i^{48}+i^{46}\right] \)
\( =i^{2}\left[\left(i^{2}\right)^{50}+\left(i^{2}\right)^{25}+\left(i^{2}\right)^{24}+\left(i^{2}\right)^{23}\right] \)
\( =(-1)\left[(-1)^{50}+(-1)^{25}+(-1)^{24}+(-1)^{23}\right] \)
 = -1[1-1+1-1] 
 = -1(0) = 0

\( =\frac{i^{10}\left[i^{582}+i^{580}+i^{578}+i^{576}+i^{574}\right]}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1 \\ \)
\( =i^{10}-1=\left(i^{4}\right)^{2} i^{2}-1 \)
 = -1-1 = -2

Given p+iq = \(\frac { (a+i)^{ 2 } }{ 2a-i } \) ...........(1)
Taking conjugate both sides we get,
p-iq =\(\frac { (a+i)^{ 2 } }{ 2a+i } \) ..........(2)
Multiplying (1) and (2) we get,
(p+iq)(p-iq) = \(\frac { (a+i)^{ 2 } }{ 2a-i } \times \frac { (a-i)^{ 2 } }{ 2a+i } \)
p²+q² = \(\frac { [(a+i)(a-i)^{ 2 }] }{ 4a^{ 2 }+1 } =\frac { ({ a }^{ 2 }+1)^{ 2 } }{ 4a^{ 2 }+1 } \).
Hence proved

Let z = x+iy ⇒ \(\bar { z } \) = x+iy
∴ \(\\ \frac { \bar { z } +1 }{ z-1 } =\frac { z-iy+1 }{ x-iy-i } =\frac { (x+1)iy }{ x-i(y+1) }\)
= \(\frac { (x+1)-iy }{ x-i(y+1) } \times \frac { x+i(y+1) }{ x+i(y+1) } \)
Choosing the real part alone we get,
\(\frac { x(x+1)+y(y+1) }{ { x }^{ 2 }+(y+1)^{ 2 } } \) = 0
⇒ x(x+1) + y(y+1) = 0
⇒ x²+x+y²+y = 0 which is the locus of z.

Given  |3z - 5| = 3 |z + 1
⇒ |3(x+iy)-5| = 3|x+iy+1|
⇒ |(3x-5)+3y| = 3|(x+1)+iy|
⇒ \(\sqrt { (3x-5)^{ 2 }+3^{ 2 } } =3\left[ \sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \right] \)
Squaring both sides we get,
(3x - 5)² + 9 = 9 [(x + 1)² + y²]
⇒ 9x² - 30x + 25 + 9 = 9 [x² + 2x + 1 + y²]
⇒ 48x - 16 = 0
⇒ 3x-1 = 0