\(\left| \frac { 1+i }{ 1-i } -\frac { 1-i }{ 1-i } \right| =\left| \frac { (1+i)^{ 2 }-(1-i)^{ 2 } }{ (1-i)((1+i) } \right| \)
= \(\left| \frac { 1+{ i }^{ 2 }+2i-1-{ i }^{ 2 }+2i }{ 1-{ i }^{ 2 } } \right| \)
= \(\left| \frac { 4i }{ 2 } \right| =|2i|=\sqrt { 4 } =2\)

We have, 
\(x+i y=\frac{(a+i b)(a+i b)}{(a-i b)(a+i b)}=\frac{a^{2}-b^{2}+2 a b i}{a^{2}+b^{2}}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}+\frac{2 a b}{a^{2}+b^{2}} i\)
So that, \(x-i y=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2 a b}{a^{2}+b^{2}} i\)
Therefore,
\(x^{2}+y^{2}=(x+i y)(x-i y)=\frac{\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}+\frac{4 a^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}=1\)

We have, \(\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}\)
\( =\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i}\)
\( =\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i\)
Therefore, conjugate of \(\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}\) is \(\frac{63}{25}+\frac{16}{25} i\).

\(\text { We have, }\left(\sqrt{1^{2}+(-1)^{2}}\right)^{x}=2^{x} \)
\( \Rightarrow \left(\sqrt{1^2+(-1)^2}\right)^{\mathrm{x}}=2^{\mathrm{x}} \ldots \ldots \ldots \ldots|\mathrm{z}|=\sqrt{\mathrm{x}^2+\mathrm{y}^2} \)
\( \Rightarrow (\sqrt{2})^{\mathrm{x}}=2^{\mathrm{x}} \)
\( \Rightarrow 2^{\frac{x}{2}}=2^{\mathrm{x}} \)
\( \Rightarrow \frac{\mathrm{x}}{2}=\mathrm{x} \)
\( \Rightarrow \mathrm{x}=2 \mathrm{x} \)
\( \Rightarrow \mathrm{x}=0 \)
Thus, x=0 is the only integral solution of the given equation. 
Therefore, the number of non-zero integral solutions of the given equation is 0.

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

\(Given, { z }_{ 1 }=3+5i\quad and\quad { z }_{ 2 }=2-3i\)
\(Now, \frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { 3+5i }{ 2-3i } =\frac { 3+5i }{ 2-3i } \times \frac { 3+5i }{ 2-3i } \)
\([by\quad rationalising\quad the\quad denominator]\)
\(=\frac { 6+9i+10i+15{ i }^{ 2 } }{ 4-9{ i }^{ 2 } } =\frac { 6+19i-15 }{ 4+9 } \quad [\because { i }^{ 2 }=-1]\)
\(=\frac { -9+19i }{ 13 } =\frac { -9 }{ 13 } +\frac { 9 }{ 13 } i\ \quad .....(i)\)
\(\because LHS=\overset { \_ \_ \_ \_ }{ \left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) } =\overset { \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ }{ \left( \frac { -9 }{ 13 } +\frac { 9 }{ 13 } i \right) } =\frac { -9 }{ 13 } -\frac { 9 }{ 13 } i\)
\(Now, consider\quad RHS =\frac { \overset { \_ \_ }{ { z }_{ 1 } } }{ \overset { \_ \_ }{ { z }_{ 2 } } } =\frac { \overset { \_ \_ \_ \_ \_ }{ 3+5i } }{ \overset { \_ \_ \_ \_ \_ \_ }{ 2-3i } } =\frac { 3+5i }{ 2-3i } \)
\(=\frac { 3+5i }{ 2-3i } \times \frac { 2-3i }{ 2-3i } \)
\([by\quad rationalising\quad denominator]\)
\(=\frac { 6-9i-10i+15{ i }^{ 2 } }{ 4-9{ i }^{ 2 } } =\frac { 6-19i-15 }{ 4+9 } \quad [\because { i }^{ 2 }=-1]\)
\(=\frac { -9-19i }{ 13 } =\frac { -9 }{ 13 } -\frac { 9 }{ 13 } i\quad .....(ii)\)
\(From\quad Eqs.(i)\quad and\quad (ii),we\quad get\)
\(\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =\frac { \overset { \_ }{ { z }_{ 1 } } }{ \overset { \_ }{ { z }_{ 2 } } } \)

Let z= \(-\sqrt { 3 } -i\) =r( \(cos\theta +isin\theta \) )
On equating real and imaginary parts, we get
\(rcos\theta =-\sqrt { 3 } ...(i)\)
\( rsin\theta =-1 ...\quad (ii)\)
On squaring and adding Eqs. (i) and (ii), we get
\({ r }^{ 2 }{ cos }^{ 2 }\theta +{ r }^{ 2 }{ sin }^{ 2 }\theta =(-\sqrt { 3 } { ) }^{ 2 }+(-1{ ) }^{ 2 }\)
\(\Rightarrow { r }^{ 2 }{ r }^{ 2 }{ cos }^{ 2 }\theta +{ r }^{ 2 }{ sin }^{ 2 }\theta )=3+1\)
\(\Rightarrow { r }^{ 2 }=4\quad [\because { r }^{ 2 }{ cos }^{ 2 }\theta +{ r }^{ 2 }{ sin }^{ 2 }\theta =1]\)
\(\Rightarrow r=2\quad \quad [\because \ r>0]\)
Now, from Eqs. (i) and (ii), we get
\(cos\theta =-\frac { \sqrt { 3 } }{ 2 } and\ sin\theta =-\frac { 1 }{ 2 } \)
\(\Rightarrow tan\theta =\frac { 1 }{ \sqrt { 3 } } \quad [\because tan\theta =\frac { sin\theta }{ cos\theta } ]\)
Now, let
\(tan\alpha =\left| tan\theta \right| ,where\quad 0<\alpha >\frac { \pi }{ 2 } \)
\(\Rightarrow \quad tan\alpha =\frac { 1 }{ \sqrt { 3 } } =tan\frac { \pi }{ 6 } \Rightarrow \alpha =\frac { \pi }{ 6 } \)
Since, z lies in III quadrant.
\(\therefore \ \theta =-\pi +\alpha\)
\(=-\pi +\frac { \pi }{ 6 } =-\frac { 5\pi }{ 6 } \)
So, polar form of z is 2( \(cos(\frac { -5\pi }{ 6 } )+isin(\frac { -5\pi }{ 6 } )\) ).
Hence, modulus of z is 2 and argument of z is \(-\frac { 5\pi }{ 6 } \)

Consider, \(\frac { 1+i }{ 1-i } =\frac { 1+i }{ 1-i } \times \frac { 1+i }{ 1+i } \)
[by rationalising the denominator]
\(=\frac { { \left( 1+i \right) }^{ 2 } }{ 1-{ i }^{ 2 } } =\frac { 1+{ i }^{ 2 }+2i }{ 1+1 }\)
\(\Rightarrow \ \frac { 1+i }{ 1-i } =\frac { 1-1+2i }{ 2 } =i\quad \left[ \because \ { i }^{ 2 }=-1 \right] ...(i)\)
\( Now,\ \frac { 1-i }{ 1+i } =\frac { 1 }{ \left( \frac { 1+i }{ 1-i } \right) } =\frac { 1 }{ i } \)
\(=\frac { 1 }{ i } \times \frac { i }{ i } =\frac { i }{ { i }^{ 2 } } =\frac { i }{ (-1) } =-i\quad \left[ \because \ { i }^{ 2 }=-1 \right] ...(ii)\)
\(Hence,\ { \left( \frac { 1+i }{ 1-i } \right) }^{ 3 }-{ \left( \frac { 1-i }{ 1+i } \right) }^{ 3 }={ i }^{ 3 }-{ \left( -i \right) }^{ 3 }\)
\(={ i }^{ 3 }+{ i }^{ 3 }=2{ i }^{ 3 }=2(-i)=0-2i\ \left[ \because \ { i }^{ 3 }=-i \right] \)
\(\therefore \ x+iy=0-2i\)
On comparing real and imaginary parts on both sides, we get x=0 and y=-2
\(\therefore \)  (x,y) = (0,-2)

We have, (5 – 3i)³ = 5³ – 3 × 5² × (3i) + 3 × 5 (3i)² – (3i)³
= 125 – 225i – 135 + 27i = – 10 – 198i.

We have,
4x + i (3x – y) = 3 + i (–6)   ... (1)
Equating the real and the imaginary parts of (1), we get
4x = 3, 3x – y = – 6,
which, on solving simultaneously, give \(x=\frac{3}{4}\) and \(y=\frac{33}{4}\)

∴ \(\left| \frac { { z }_{ 1 }+{ z }_{ 2 }+1 }{ { z }_{ 1 }-{ z }_{ 2 }+1 } \right| \)
= \(\left| \frac { 2-i+1+i+1 }{ 2-i-1-i+1 } \right| =\left| \frac { 4 }{ 2-2i } \right| \)
= \(\frac { 4 }{ |2-2i| } =\frac { 4 }{ \sqrt { (2)^{ 2 }+(-2)^{ 2 } } } \)
= \(\frac { 4 }{ \sqrt { 4+4 } } =\frac { 4 }{ \sqrt { 8 } } =\frac { 4 }{ 2\sqrt { 2 } } =\sqrt { 2 } \)

Multiplicative inverse of -i
= \(\frac { 1 }{ -i } =\frac { i }{ -{ i }^{ 2 } } =\frac { 1 }{ -(-1) } =i\)

M.I of  \(\sqrt { 5 } \)+3i = \(\frac { 1 }{ \sqrt { 5 } +3i } \)
= \(\frac { 1 }{ \sqrt { 5 } +3i } \times \frac { \sqrt { 5 } -3i }{ \sqrt { 5 } -3i } \)
= \(\frac { \sqrt { 5 } -3i }{ (\sqrt { 5 } )^{ 2 }+(3i)^{ 2 } } =\frac { \sqrt { 5 } -3i }{ 5-9i^{ 2 } } \)
= \(\frac { \sqrt { 5 } -3i }{ 5+9 } \)
\(=\frac{\sqrt{5}}{14}-\frac{3 i}{14}\)

Multiplicative inverse of  (4-3i) = \(\frac { 1 }{ 4-3i } =\frac { 1 }{ 4-3i } \times \frac { 4+3i }{ 4+3i } \)
= \(\frac { 4+3i }{ (4)^{ 2 }+(3i)^{ 2 } } =\frac { 4+3i }{ 16-9i^{ 2 } } \)
= \(\frac { 4+3i }{ 16+9 } \)
= \(\frac{4}{25}+\frac{3}{25} i\)