(-3i) (i)\(\left( -\frac { 1 }{ 4 } i \right) ^{ 3 }\)
= -3i²\(\times -\frac { 1 }{ 64 } { i }^{ 3 }=\frac { 3 }{ 64 } { i }^{ 5 }\)
= \(\frac { 3 }{ 64 } \)(i²)².i=\(\frac { 3 }{ 64 } \)i= 0+ \(\frac { 3 }{ 64 } \)i

\(\frac { 4 }{ 17 } \pm \frac { 1 }{ 17 } i\)

\(\frac{3}{5}\pm\frac{1}{5}i\)

\(|z|=\sqrt 2,arg(z)=\frac{3\pi}{4};\sqrt 2(cos\frac{3\pi}{4}+isin\frac{3\pi}{4})\)

sec∝(cos∝+i sin∝)

\(\frac { 1 }{ 49 } -\frac { 4\sqrt { 3 } }{ 49 } i\)

4

8i

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

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

\(\left( -2-5i \right) \div (3-6i)=\frac { -2-5i }{ 3-6i } \)
\(=\frac { -(2+5i) }{ 3-6i } \times \frac { (3+6i) }{ (3+6i) } \)
[ by rationalizing the denominator]
\(= -\frac { \left[ 6+12i+15i+30{ i }^{ 2 } \right] }{ { 3 }^{ 2 }-{ \left( 6i \right) }^{ 2 } } \left[ \because \ ({ z }_{ 1 }+{ z }_{ 2 })\quad ({ z }_{ 1 }-{ z }_{ 2 })={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } \right] \)
\(=\frac { -\left[ 6+27i-30 \right] }{ 9+36 } \quad \left[ \because \ { i }^{ 2 }=-1 \right] \)
\(=\frac { -\left( -24+27i \right) }{ 45 } =\frac { 24 }{ 45 } -\frac { 27 }{ 45 } i\)
\(=\frac { 8 }{ 15 } -\frac { 3 }{ 5 } i=\frac { 8 }{ 5 } +i\left( \frac { -3 }{ 5 } \right) \)
Which is in the form (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.

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, \({ 2x }^{ 4 }+{ 5x }^{ 3 }+{ 7x }^{ 2 }-x+41\)
\(=\left( { x }^{ 2 }+4x+7 \right) ({ 2x }^{ 2 }-3x+5)+6\)
\([\because dividend=quotient\times divisor+remainder]\)
\(=0\times (({ 2x }^{ 2 }-3x+5)+6=6\quad [\because { x }^{ 2 }+4x+7=0]\)

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

\(|{ z }_{ 1 }-{ z }_{ 2 }|=|2-3i|=\sqrt { 4+9 } =\sqrt { 13 } \)
\( |{ z }_{ 1 }|=\sqrt { 9+1 } =\sqrt { 10 } \) 
and \(|{ z }_{ 2 }|=\sqrt { 1+16 } =\sqrt { 17 }\)
 

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 
\(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } +i\quad sin\frac { 3\pi }{ 4 } \right) \)

\(z=\frac { 2+3i }{ 3+2i } x \frac { 3-2i }{ 3-2i } =\frac { 12+5i }{ 13 } \)
\(\overline { z } =\frac { 12 }{ 13 } -\frac { 5 }{ 13 } i\quad and\quad \left| z \right| =1\)

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 \(\frac { b }{ a } =\frac { \frac { 2x }{ { x }^{ 2 }+1 } }{ \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } } =\frac { 2x }{ { x }^{ 2 }-1 } \)

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
\(\therefore\sqrt {-5+12i}=\pm(2+3i)\)

\(\frac { 1+3i }{ 1-2i } \times \frac { 1+2i }{ 1+2i } =\frac { 1+2i+3i+6{ i }^{ 2 } }{ 1-4{ i }^{ 2 } } \)
=\(\frac { -5+5i }{ 5 } \)=-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 \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } +isin\frac { 3\pi }{ 4 } \right) \)