Complex Numbers

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

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.

The modulus of the complex number $ z $ such that $ |z+3-i|=1 $ and $ \arg (z)=\pi $ is equal to

The points represented by the complex number $ 1+i,-2+3 i, \frac{5}{3} i $ on the argand plane are

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

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

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

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

If \(z = r{e^{i\theta }}\), then arg(eiz) 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 \(z = {{7 + i} \over {3 + 4i}}\), then z14 is