Question

If z be a non- real complex number satisfying |z|=2, then which of the following is/are true?

• A. $arg\left(\frac{z-2}{z+2}\right)=\pm \frac{\pi }{2}$
• B. $arg\left(\frac{z+1+i\sqrt{3}}{z-1+i\sqrt{3}}\right)=\frac{\pi }{6}$
• C. $|z^2-1|\ge 3$
• D. $|z^2-1|\le 5$

Answer 1

Answer: (A), (C), (D)

The correct options are
A

$arg\left(\frac{z-2}{z+2}\right)=\pm \frac{\pi }{2}$

C

$|z^2-1|\ge 3$

D

$|z^2-1|\le 5$

(a) For P(z), arg $\left(\frac{z-2}{z+2}\right)=\frac{\pi }{2}$

and for Q(z), arg $\left(\frac{z-2}{z+2}\right)=\frac{-\pi }{2}$

$\Rightarrow$ (a) is true.

(b) $\Delta$ AOB is equilateral.

$\therefore \angle AOB=\frac{\pi }{3}$ and $\angle ACB=\frac{\pi }{6}$

$\therefore arg\left(\frac{z-1+i\sqrt{3}}{z+1+i\sqrt{3}}\right)=\frac{\pi }{6}\left(Byrotation\right)$

$\Rightarrow arg\left(\frac{z+1+i\sqrt{3}}{z-1+i\sqrt{3}}\right)=-\frac{\pi }{6}$

$\Rightarrow \left(b\right)$ is not true.

(c) and (d)

We have $|z^2-1{|}^2=(z^2-1)({\overline{z}}^2-1)=z^2{\overline{z}}^2-z^2-{\overline{z}}^2+1$

Taking $z=x+iy,$

$|z^2-1{|}^2=|z{|}^4-2(x^2-y^2)+1=17-2(x^2-y^2)=25-4x^2.........\because x^2+y^2=4$

Now given |z| = 2 $\Rightarrow \text{range of}x\text{is}[-2,2]$

$\Rightarrow 9\le |z^2-1{|}^2\le 25$

Hence, $3\le |z^2-1|\le 5$

$\Rightarrow$ (c) and (d) are true.