Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 1: TS EAMCET - 5 August 2021 Shift 1

If $z_1=1-2i, z_2=1+i$ and $z_3=3+4i,$ then $\left|\left(\frac{1}{z_1}+\frac{2}{z_2}\right)\frac{z_3}{z_2}\right|=$

If $2+2\sqrt{3}i=k(\cos \theta +i\sin \theta ), (k>0),$ then $\frac{1}{\sqrt{3}}[\cos 6\theta +i\sin 6\theta ]=$

If the roots of the equation $(z-4{)}^3=8i$ are $a-2i, b+i,$ and $c+i,$ then $\sqrt{abc}=$

If $x=\frac{4}{5}+\frac{3}{5}i, y=\frac{\sqrt{3}}{\sqrt{8}}-\frac{\sqrt{5}}{\sqrt{8}}i,$ then $\left(x^2+\frac{1}{x^2}\right)\left(y^2-\frac{1}{y^2}\right)=$

For $n\in N,{\left(\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta }\right)}^n=$

$\omega$ is a complex cube root of unity. Match the items of List-I to the items of List-II

If $\omega$ represents a cube root of unity and $\sum _{k=1}^n\left(k+\frac{1}{\omega }\right)\left(k+\frac{1}{{\omega }^2}\right)=340$, then $n=$

If $\alpha , \beta$ are the roots of the equation $1+x+x^2=0,$ then $(2-\alpha )(2-\beta )\left(2-{\alpha }^{10}\right)\left(2-{\alpha }^{20}\right)=$