If $|{z_1}|\, = \,|{z_2}|$ and $arg\,\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $, then ${z_1} + {z_2}$ is equal to
$0$
Purely imaginary
Purely real
None of these
If $\frac{\pi }{2} < \alpha < \frac{3}{2}\pi $ , then the modulus and argument of $(1 + cos\, 2\alpha ) + i\, sin\, 2\alpha $ is respectively
Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0, z \in C$. Then $4\left(\alpha^2+\beta^2\right)$ is equal to :
Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation
$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$ is. . . . . .
Let $z,w$be complex numbers such that $\overline z + i\overline w = 0$and $arg\,\,zw = \pi $. Then arg z equals
${\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2}$ is equal to