Question

Match the elements of List 1 with elements of List 2

Answer 1

(A) $||z|-\frac{1}{|z|}|\le |z+\frac{1}{z}|$
$-2\le |z|-\frac{1}{|z|}\le 2$
$|z{|}^2+2|z|-1\ge 0$ and $|z{|}^2-2|z|-1\le 0$
$|z|\ge \sqrt{2}-1,|z|\le \sqrt{2}+1$
$|z{|}_{min}=\sqrt{2}-1$
so minimum value of $\frac{|z|}{tan\frac{\pi }{8}}=1$
(B) $|z|=1$
Let $z=cos\theta +isin\theta$
$\frac{z^n}{z^{2n}+1}-\frac{z^{-n}}{(\overline{z}{)}^{2n}+1}$
$=\frac{cosn\theta +isinn\theta }{1+cos2n\theta +isin2n\theta }-\frac{cosn\theta -isinn\theta }{1+cos2n\theta -isin2n\theta }$
$=\frac{cosn\theta +isinn\theta }{2cosn\theta (cosn\theta +isinn\theta )}-\frac{cosn\theta -isinn\theta }{2cosn\theta (cosn\theta -isinn\theta )}$
$=\frac{1}{2cosn\theta }-\frac{1}{2cosn\theta }$
$=0$
(C) $8i{z}^3+12z^2-18z+27i=0$
$\Rightarrow (2iz+3)(4z^2+9i)=0$
$\Rightarrow z=\frac{3}{2}i,z^2=-\frac{9}{4}i\Rightarrow 2|z|=3$
(D) $z^4+z^2+z^2+z+1$
$=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
Put $z=-2$
${\Pi }_{i=1}^4(z_i+2)=(-2{)}^4+(-2{)}^3+(-2{)}^2+(-2)+1=11$