\(\frac { \left( 3+i\sqrt { 5 } \right) \left( 3-i\sqrt { 5 } \right) }{ \left( \sqrt { 3 } +i\sqrt { 2 } \right) -\left( \sqrt { 3 } -i\sqrt { 2 } \right) } =\frac { 9+5 }{ 2i\sqrt { 2 } } =\frac { 14 }{ 2\sqrt { 2 } } \times \frac { i }{ { i }^{ 2 } } =\frac { -14i }{ 2\sqrt { 2 } } \)
\(Ans.-\frac { 7\sqrt { 2 } }{ 2 } i\)

\(Given,\ { 2x }^{ 2 }-3ix+2=0\)
On comparing Eq.(i) with ax²+bx+c=0, we get a = 2, b = 3i and c = 2
\( \because \quad \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } and\ \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a }\)
\( \therefore \quad \alpha =\frac { -3i+\sqrt { { (3i) }^{ 2 }-4\times 2\times 2 } }{ 2\times 2 } =\frac { -3i+\sqrt { -9-16 } }{ 4 }\)
\(=\frac { -3i+\sqrt { -25 } }{ 4 } =\frac { -3i+5i }{ 4 } \)
\(\Rightarrow \beta =\frac { -3i-\sqrt { { (3i) }^{ 2 }-4\times 2\times 2 } }{ 2\times 2 } =\frac { -3i-\sqrt { -9-16 } }{ 4 } \)
\( \Rightarrow =\frac { -3i+\sqrt { -25 } }{ 4 } =\frac { -3i-5i }{ 4 } \)

Multiply numerator and denominator by \(1+\sqrt { 2 } i\), we get
\(\frac { 5+\sqrt { 2 } i }{ 1-\sqrt { 2 } i } \times \frac { 1+\sqrt { 2 } i }{ 1+\sqrt { 2 } i } =\frac { 3+6\sqrt { 2 } i }{ 1+2 } \)
\(\text {Ans} : 1+2\sqrt { 2 } i\)

\(z=\frac { 1+icos\theta }{ 1-2icos\theta } \times \frac { 1+2icos\theta }{ 1+2icos\theta } =\frac { 1-2{ cos }^{ 2 }\theta +3icos\theta }{ 1+4{ cos }^{ 2 }\theta } \)
\(For\quad Re(z),put\quad 3cos\theta =0\Rightarrow cos\theta =0\)
Ans. \(2n\pi \pm \frac { \pi }{ 2 } \)

\(We\ have,\ 2{ x }^{ 2 }-2\sqrt { 3x } +\frac { 21 }{ 8 } =0\)
On comparing Eq.(i) with ax²+bx+c=0, we get
\(a=2,\ b=-2\sqrt { 3 } \ and\ c=\frac { 21 }{ 8 } \)
\(\because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \)
\(\therefore x=\frac { -(-2\sqrt { 3 } \pm \sqrt { { (-2\sqrt { 3 } ) }^{ 2 }-4\times 2\times \frac { 21 }{ 8 } } }{ 2\times 2 } \)
\(=\frac { 2\sqrt { 3 } \pm \sqrt { 12-12 } }{ 4 } =\frac { 2\sqrt { 3 } \pm \sqrt { -9 } }{ 4 }\)
\(=\frac { 2\sqrt { 3 } \pm 3i }{ 4 } =\frac { \sqrt { 3 } }{ 2 } \pm \frac { 3 }{ 4 } i. \quad [\because \sqrt { -1 } =i]\)
\(\text{Hence, the roots are} \frac { \sqrt { 3 } }{ 2 } +\frac { 3 }{ 4 } i\ and \ \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i.\)

Let z = (3 - 2i) (3 + 2i) (1 + i)
z = (9 = 6i - 6i - 4i²) (1 + i)
= ( 9 + 4) (1 + i) = 13 + 13i
\(\overline { z } =13-13i\ and\ \left| z \right| =13\sqrt { 2 } \) 

\(z=\frac { 1-i }{ 1+i } x \frac { 1-i }{ 1-i } =\frac { 1-1-2i }{ 1+1 } =-i\)   = i

Write the complex number in the form \(\frac { a+ib }{ c+id } \) and then rationalising the denominator, further simplify it
\(\frac { \left( 3+\sqrt { 5i } \right) \left( 3-\sqrt { 5i } \right) }{ \left( \sqrt { 3 } +\sqrt { 2i } \right) -\left( \sqrt { 3 } -\sqrt { 2i } \right) } \)
\(=\frac { { \left( 3 \right) }^{ 2 }-{ \left( \sqrt { 5i } \right) }^{ 2 } }{ \sqrt { 3 } +\sqrt { 2i } -\sqrt { 3 } +\sqrt { 2i } } \quad \left[ \because \ \left( { z }_{ 1 }+{ z }_{ 2 } \right) \left( { z }_{ 1 }-{ z }_{ 2 } \right) ={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } \right] \)

\(=\frac { 9+5 }{ 2\sqrt { 2i } } =\frac { 14 }{ 2\sqrt { 2i } } =\frac { 7 }{ \sqrt { 2i } } \times \frac { \sqrt { 2i } }{ \sqrt { 2i } } \)
[by rationalising the denominator]
\(=\frac { 7\sqrt { 2i } }{ 2{ i }^{ 2 } } =\frac { 7\sqrt { 2i } }{ -2 } =0-i\frac { 7\sqrt { 2 } }{ 2 } \)
\(=0+i\left( \frac { -7\sqrt { 2 } }{ 2 } \right) \)
Which is in the form of (a+ib).

Consider the given expression.
\(\left[ \left( \frac { 1 }{ 3 } +\frac { 7 }{ 3 } i \right) +\left( 4+\frac { 1 }{ 3 } i \right) \right] -\left( -\frac { 4 }{ 3 } +i \right) \)
\(=\left[ \left( \frac { 1 }{ 3 } +4 \right) +i\left( \frac { 7 }{ 3 } +\frac { 1 }{ 3 } \right) \right] -\left( -\frac { 4 }{ 3 } +i \right) \)
\(=\left( \frac { 13 }{ 3 } +\frac { 8 }{ 3 } i \right) +\left( \frac { 4 }{ 3 } -i \right) =\left( \frac { 13 }{ 3 } +\frac { 4 }{ 3 } \right) +i\left( \frac { 8 }{ 3 } -1 \right) \) 
 \(=\frac { 17 }{ 3 } +\frac { 5 }{ 3 } i\), which is in the form of a + ib.

Consider the given expression,
\(\frac { { i }^{ 592 }+{ i }^{ 590 }+{ i }^{ 588 }+{ i }^{ 586 }+{ i }^{ 584 } }{ { i }^{ 582 }+{ i }^{ 580 }+{ i }^{ 578 }+{ i }^{ 576 }+{ i }^{ 574 } } \)
\(\frac { { i }^{ 584+8 }+{ i }^{ 584+6 }+{ i }^{ 584+4 }+{ i }^{ 584+2 }+{ i }^{ 584 } }{ { i }^{ 574+8 }+{ i }^{ 574+6 }+{ i }^{ 574+4 }+{ i }^{ 574+2 }+{ i }^{ 574 } } \)
\(=\frac { { i }^{ 584 }({ i }^{ 8 }+{ i }^{ 6 }+{ i }^{ 4 }+{ i }^{ 2 }+1) }{ { i }^{ 574 }({ i }^{ 8 }+{ i }^{ 6 }+{ i }^{ 4 }+{ i }^{ 2 }+1) } \)
\(=\frac { { i }^{ 584 } }{ { i }^{ 574 } } ={ i }^{ 584-574 }={ i }^{ 10 }\)
= i^4x2+2 = (i⁴)² . i²
= (1)² . i² = -1   [i⁴ =1 and i² =-1]

We have. \(x=-2-\sqrt { 3i } \)
\(\Rightarrow x=-2-\sqrt { 3i } \)
On squaring both sides, we get
\((x+2)^{2}=(-\sqrt{3} i)^{2} \Rightarrow x^{2}+4+4 x=3 i^{2}\)
\(\left[\because\left(z_{1}+z_{2}\right)^{2}=z_{1}^{2}+z_{2}^{2}+2 z_{1} z_{2}\right]\)
\(\Rightarrow \ x^{2}+4 x+4=-3 \quad\left[\because i^{2}=-1\right]\)
\(\Rightarrow \ x^{2}+4 x+7=0\)
Now divide \(2 x^{4}+5 x^{3}+7 x^{2}-x+41 \text { by } x^{2}+4 x+7\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 x^{2}-3 x+5 \\ x ^ { 2 } + 4 x + 7 \sqrt { 2 x ^ { 4 } + 5 x ^ { 3 } + 7 x ^ { 2 } - x + 4 1 } \)
\(\begin{aligned} 2 x^{4}+8 x^{3}+14 x^{2} \\ \frac{- \ -}{-3 x^{3}-7 x^{2}-x+41} \end{aligned}\)
\(\begin{aligned} 3 x^{3}-12 x^{2}-21 x^{} \\ \frac{+ \ + \ +}{5 x^{2}+20 x+35} \end{aligned}\)
\(\begin{aligned} {5 x^{2}+20 x+35} \\ \frac{- \ - \ -} {6} \end{aligned}\)
Thus, \({ 2x }^{ 4 }+{ 5x }^{ 3 }+{ 7x }^{ 2 }-x+41\)
\(=\left( { x }^{ 2 }+4x+7 \right) ({ 2x }^{ 2 }-3x+5)+6\)
\([\because dividend=quotient\times divisor+remainder]\)
\(=0\times (({ 2x }^{ 2 }-3x+5)+6=6\quad [\because { x }^{ 2 }+4x+7=0]\)

Given \(\frac { \left( 1+i \right) x-2i }{ 3+i } +\frac { \left( 2-3i \right) y+i }{ 3-i } =i\)
\(\Rightarrow \frac { x+\left( x-2 \right) i }{ 3+i } +\frac { 2y+\left( 1-3y \right) i }{ 3-i } =i\)
\(\Rightarrow \frac { \left[ x+\left( x-2 \right) i \right] \left( 3-i \right) +\left[ 2y+\left( 1-3y \right) i \right] \left( 3+i \right) }{ \left( 3+i \right) \left( 3-i \right) } =i\)
\(\Rightarrow \left( 4x+9y-3 \right) +i\left( 2x-7y-3 \right) =10i\)
\(\Rightarrow 4x+9y-3=0\quad and\quad 2x-7y-3=10\)
\(Ans.\ x=3\ and \ y=-1\)

-160

Let z = x + iy, then
\(\frac { z-1 }{ z+1 } =\frac { ({ x }^{ 2 }-1)+{ y }^{ 2 }+i[y(x+1)-y(x-1)] }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } \\ \)
\(\because \frac { z-1 }{ z+1 } \)  is purely imaginary.
\(\therefore Re(\frac { z-1 }{ z+1 } ) =0\ i.e.\ \frac { ({ x }^{ 2 }-1)+{ y }^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } =0\)
\({ x }^{ 2 }-1-{ y }^{ 2 }=0\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1= |z|=1\)

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.