Complex Numbers...

If $$z_1, z_2$$ be two non zero complex numbers satisfying the equation $$\displaystyle \left | \frac{z_1 + z_2}{z_1 - z_2} \right | = 1$$ then $$\displaystyle \frac{z_1}{z_2} + \left ( \frac{z_1}{z_2} \right )$$ is

Find the range of real number $$\alpha$$ for which the equation $$z + \alpha |z - 1| + 2i = 0;  z= x + iy$$ has a solution. Find the solution.

If z be a complex number satisfying$$\displaystyle\ z^{4}+z^{3}+2z^{2}+z+1=0$$ then $$\displaystyle\ |z|$$ is 

If $$z = x+iy$$ and $$w = \dfrac{(1-iz)}{(z-i)}$$, then $$|w| = 1$$ implies that, in the complex plane

$$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 }  \right)  }^{ 6 }+{ \left( \frac { i-\sqrt { 3 }  }{ 2 }  \right)  }^{ 6 }=$$

Let $$\displaystyle\ z_{1}= a+ib, z_{2}= p+iq$$ be two unimodular complex numbers such that $$\displaystyle\ Im(z_{1}z_{2})=1$$. If$$\displaystyle\ \omega_{1}= a+ip, \omega_{2}=b+iq$$ then

The root of $$(x + a) (x + b) - 8k = (k - 2)^2$$ are real and equal, when $$a,b,c$$ $$\epsilon$$ R, then

If $$\displaystyle y^{2}< x$$ and $$\displaystyle x\in \left ( -\infty ,0 \right )$$  then $$y$$ must