Complex Numbers · Mathematics · BITSAT

If ' $a$ ' is a complex number such that $|a|=1$. Find the value of $a$, so that the equation $a z^2+z+1=0$ has one purely imaginary root.

Let $z$ be a complex number for which $\left|2 z \cos \theta+z^2\right|>1$, if $|z|

Number of solutions of the equation $$z^2+|z|^2=0$$ and $$z \neq 0$$ is

If $$z_1$$ and $$z_2$$ be nth root of unity which subtend a right angled at the origin. Then, $$n$$ must be of the form

If $$|w| = 2$$, then the set of points $$z = w - {1 \over w}$$ is contained in or equal to the set of points z satisfying

The smallest positive integral value of n such that $${\left[ {{{1 + \sin {\pi \over 8} + i\cos {\pi \over 8}} \over {1 + \sin {\pi \over 8} - i\cos {\pi \over 8}}}} \right]^n}$$ is purely imaginary, is equal to

If Re(z + 2) = | z $$-$$ 2 |, then the locus of z is

If $$z = {{7 + i} \over {3 + 4i}}$$, then z¹⁴ is

The root of the equation $$2(1 + i){x^2} - 4(2 - i)x - 5 - 3i = 0$$, where $$i = \sqrt { - 1} $$, which has greater modulus, is

If $$z = r{e^{i\theta }}$$, then arg(e^iz) is