Given $z$ is a complex number such that $|z| < 2,$ then the maximum value of $|iz + 6 -8i|$ is equal to-
$10$
$8$
$12$
$6$
$|i z+6-8 i| \leq|i z|+|6-8 i|=2+10=12$.
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. . . . . .
If $\frac{{2{z_1}}}{{3{z_2}}}$ is a purely imaginary number, then $\left| {\frac{{{z_1} – {z_2}}}{{{z_1} + {z_2}}}} \right|$ =
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
The conjugate of the complex number $\frac{{2 + 5i}}{{4 – 3i}}$ is
If $z = x + iy\, (x, y \in R,\, x \neq \, -1/2)$ , the number of values of $z$ satisfying ${\left| z \right|^n}\, = \,{z^2}{\left| z \right|^{n – 2}}\, + \,z{\left| z \right|^{n – 2}}\, + \,1\,.\,\left( {n \in N,n > 1} \right)$ is
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