Complex Numbers · Mathematics · TS EAMCET

If $\omega$ is the complex cube root of unity and

$\left(\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\right)^{k}+\left(\frac{a+b \omega+c \omega^{2}}{b+a \omega^{2}+c \omega}\right)^{l}=2$, then $2 k+l$ is always

If $\sqrt{5}-i \sqrt{15} \doteqdot r(\cos \theta+i \sin \theta),-\pi<\theta<\pi$, then $r^2\left(\sec \theta+3 \operatorname{cosec}^2 \theta\right)=$

The point $P$ denotes the complex number $z=x+i y$ in the argand plane. If $\frac{2 z-i}{z-2}$ is a purely real number, then the equation of the locus of $P$ is

$x$ and $y$ are two complex numbers such that $|x|=|y|=1$.

If $\arg (x)=2 \alpha, \arg (y)=3 \beta$ and $\alpha+\beta=\frac{\pi}{36}$, then $x^6 y^4+\frac{1}{x^6 y^4}=$

$\operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right)=$

$$ \text { If } x+i y=\sqrt{\frac{3+i}{1+3 i}}, \text { then }\left(x^2+y^2\right)^2= $$

If the imaginary part of $\frac{2 z+1}{i z+1}$ is -2, then the locus of the point representing $z$ in the Argand plane is

If $i=\sqrt{-1}$, then $(1+i)^{10}+(1-i)^{10}=$