Question

Match the following

Column - I	Column - II
(A) If z is a complex number satisfying $|z^3+z^{-3}|\le 2m\text{then}|z+\frac{1}{z}|$ can't take the value	(p) 4
(B) If z is a complex number satisfying $|z^3+z^{-3}|\le 2,\text{then}|z+\frac{1}{z}|$ can take the value	(q) 1
(C) $\left|{[i^{19}+\frac{1}{i^{25}}]}^2\right|$ equals	(r) 2
(D) ${\left(\frac{1+i}{1-i}\right)}^{2008}$ equals	(s) 5
	(t) 3

• A.
• B. A(p, s, t),B(q, r, t),C(p),D(q)

Answer 1

The correct option is B A(p, s, t),B(q, r, t),C(p),D(q)

(A) $|z^2+\frac{1}{z^3}|\le 2$

Let $|z+\frac{1}{z}|=a$

then, $\left|{(z+\frac{1}{z})}^3\right|=|z^3+\frac{1}{z^3}+3(z+\frac{1}{z})|\le |z^3+\frac{1}{z^3}|+\left|3(z+\frac{1}{z})\right|\le 2+3a$

$\Rightarrow a^3\le 2+3a$

$\Rightarrow a^3-3a-2\le 0$

$\Rightarrow (a-2)(a+1{)}^2\le 0$

This gives $a\le 2$

(B) From the bound obtained in (A) one can answer this question

(C) ${(i^{19}+\frac{1}{i^{25}})}^2={(i^3+\frac{1}{i})}^2={\left(\frac{i^4+1}{i}\right)}^2=\frac{2^2}{i^2}=-4$

(D) ${\left(\frac{1+i}{1-i}\right)}^{2008}=i^{2008}=(i^4{)}^{502}=1$