\(\left[ \frac { 3+2i }{ 2-3i } \right] +\left[ \frac { 3-2i }{ 2+3i } \right] \)
= \(\frac { 3+2i }{ 2-3i } \times \frac { 2+3i }{ 2+3i } +\frac { 3-2i }{ 2+3i } \times \frac { 2-3i }{ 2-3i } \)
= \(\frac { 6+13i+6{ i }^{ 2 } }{ 4-{ 9i }^{ 2 } } +\frac { 6-13i+6{ i }^{ 2 } }{ 4-{ 9i }^{ 2 } } \)
= \(\frac { 6+13i+6{ i }^{ 2 }+6-13i+6{ i }^{ 2 } }{ 4-9 } \)
= \(\frac { 12+12{ i }^{ 2 } }{ 13 } =\frac { 12-12 }{ 13 } =0\)    [∵ i²=-1]
Which is purely real

\(\frac { 3+i }{ 2 } and\quad 3i\)

\(\frac{-7}{26}\pm\frac{\sqrt 3}{26}i\)

\(\frac{9}{16}\pm\frac{\sqrt {15}}{16}i\)

We have , \(\frac{3+2isin\theta}{1-2isin\theta}=\frac{(3+2i\ sin\theta)}{(1-2i\ sin\theta)}\times\frac{(1+2i\ sin\theta)}{1+2i\ sin\theta}\)
\(=\frac{(3-4\ sin\theta)+(6\ sin\theta+2\ sin\theta)i}{1-4\ sin^2\theta}=\frac{(3-4sin\theta)}{1+4sin^2\theta}+\frac{8sin\theta i}{1+4sin^2\theta}\)
Since \(\frac{3+2isin\theta}{1-2isin\theta}\) is purely real.
\(\therefore\frac{8sin\theta}{1+4sin^2\theta}=0\Rightarrow8sin\theta=0\Rightarrow sin\theta=0\Rightarrow \theta=n\pi,n \in Z\)

cos \(\theta \)-i sin \(\theta \)

\(\frac { 3 }{ 13 } -\frac { 2i }{ 13 } \)

\(\left[ \frac { 1-cos\theta }{ 2-2cos\theta +3sin^{ 2 }\theta } \right] +i\left[ \frac { -2sin\theta }{ 2-2cos\theta +3sin^{ 2 }\theta } \right] \)

8i

z = (6 + 5i )² = 36 - 25 + 60i = 11 + 60i 
= 11 - 60i

\(Put\quad z=x+iy\)
\(\therefore \quad \left| x+iy \right| =(x+iy)+1+2i\)
\(\Rightarrow \sqrt { { x }^{ 2 }+{ y }^{ 2 } } =(x+1)+i(y+2)\)
\(\Rightarrow \sqrt { { x }^{ 2 }+{ y }^{ 2 } } =x+1\quad and\quad y+2=0\)
Ans. \(\frac { 3 }{ 2 } -2i\)

\(\frac { 2-i }{ 2 } \)

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

Here x²-2x+\(\frac { 3 }{ 2 } \) =0
Comparing the given quadratic equation with ax²+bx+c =0, we have
a=1, b=-2 and =\(\frac { 3 }{ 2 } \)
∴ x= \(\frac { (-2)\pm \sqrt { (-2)^{ 2 }-4\times 1\times \frac { 3 }{ 2 } } }{ 2\times 1 } \)
= \(\frac { 2\pm \sqrt { 4-6 } }{ 2 } =\frac { 2\pm \sqrt { -2 } }{ 2 } \)
= \(\frac { 2\pm \sqrt { 2 } i }{ 2 } =1+\frac { \sqrt { 2 } }{ 2 } i\)
Thus x = \(1+\frac { \sqrt { 2 } }{ 2 } i\) and x = 1 -\(\frac { \sqrt { 2 } }{ 2 } i\)

Here z=\(\sqrt { 3 } \)+1 = r (cos\(\theta \)+i sin\(\theta \))
⇒ r cos\(\theta \) =\(\sqrt { 3 } \) and r sin\(\theta \) =1   ..(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \))=3+1
⇒ r²= 4⇒ r =2
∴  2 cos \(\theta \)=\(\sqrt { 3 } \)and 2 sin \(\theta \)=1
∴  cos\(\theta \)=\(\frac { \sqrt { 3 } }{ 2 } \)and sin\(\theta \) =\(\frac { 1 }{ 2 } \)
Since sin\(\theta \) and cos\(\theta \) are both positive
∴  \(\theta \) lies in first quadrant
∴   \(\theta \)  = \(\frac { \pi }{ 6 } \)
Hence polar form of z is 2\(\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)