Let z = x + iy, then \(|z-1|=|z+1|\)
\(\Rightarrow \left| x+iy-1 \right| =\left| x+iy+1 \right| \)
\(\Rightarrow \left| (x-1)+iy \right| =\left| (x+1)-iy \right| \)
\(\Rightarrow\sqrt { (x-1)^{ 2 }+{ y }^{ 2 } } =\sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \)
\(\Rightarrow (x-1)^{ 2 }+{ y }^{ 2 }=(x+1)^{ 2 }+{ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+1-2x={ x }^{ 2 }+1+2x\Rightarrow 4x=0\Rightarrow x=0\)
\(\therefore \quad Re(z)=0\)

\(8\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)

We have, \(\sqrt { -25 } +3\sqrt { -4 } +2\sqrt { -9 } \)
\(=\sqrt { 25 } \sqrt { -1 } +3\sqrt { 4 } \sqrt { -1 } +2\sqrt { 9 } \sqrt { -1 }\)
= 5 x i + 3 x 2 x i + 2 x 3 i
= 5i + 6i + 6i
= 17i

i

\(z=\frac { 1 }{ (1-cos\theta )+2-sin\theta } \times \frac { (1-cos\theta )-2isin\theta }{ (1-cos\theta )-2isin\theta } \)
\(=\frac { (1-cos\theta )-2isin\theta }{ { (1-cos\theta ) }^{ 2 }+4{ sin }^{ 2 }\theta } =\frac { (1-cos\theta )-2isin\theta }{ 1+{ cos }^{ 2 }\theta -2cos\theta -4{ sin }^{ 2 }\theta } \)
Ans. \((\frac { 1-cos\theta }{ 2-2cos\theta +3{ sin }^{ 2 }\theta } )+i(\frac { -2sin\theta }{ 2-2cos\theta +3{ sin }^{ 2 }\theta } )\)

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

Now \(\left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| ^{ 2 }\)
= \(\left| \frac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right| \left| \frac { \overline { \beta -\alpha } }{ \overline { 1-\overline { \alpha } \beta } } \right| \)              \(\left[ \because |z|^{ 2 }=z\overline { z } \right] \)  
= \(\left| \frac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right| \left| \frac { \overline { \beta } -\overline { \alpha } }{ \overline { 1-\alpha \overline { \beta } } } \right| \)
=  \(\frac { \beta \overline { \beta } -\beta \overline { \alpha } -\alpha \overline { \beta } +\alpha \overline { \alpha } }{ 1-\overline { \alpha } \beta -\alpha \overline { \beta } +\alpha \overline { \alpha } \beta \overline { \beta } } \)
= \(\frac { |\beta |^{ 2 }-\overline { \alpha } \beta -\alpha \overline { \beta } +|\alpha |^{ 2 } }{ 1-\overline { \alpha } \beta -\alpha \overline { \beta } +|\alpha |^{ 2 }|\beta |^{ 2 } } \)
= \(\frac { 1-\overline { \alpha } \beta -\alpha \overline { \beta } +|\alpha |^{ 2 } }{ 1-\overline { \alpha } \beta -\overline { \alpha } \beta +|\alpha |^{ 2 } } =1\)
∴\(\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|\) =1

Here z =-1-i=r(cos\(\theta \)+isin\(\theta \))
⇒ r cos =-1 and r sin = -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\(\theta \) and cos\(\theta \) are both negative
∴ \(\theta \)  lies in third quadrant
∴ \(\theta \)= \(\left( -\pi +\frac { \pi }{ 4 } \right) =\frac { -3\pi }{ 4 } \)
Hence polar form of z is
\(\sqrt { 2 } \left[ cos\left( \frac { -3\pi }{ 4 } \right) +i sin\left( \frac { -3\pi }{ 4 } \right) \right] \)