Question

Let $z$ be a complex number such that $|2z+\frac{1}{z}|=1$ and $\arg \left(z\right)=\theta$, then minimum value of $8{\sin }^2\theta$ is

• A. $0$
• B. $5$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

Given, $|2z+\frac{1}{z}|=1\cdots \left(i\right)$
We know, $z=r(\cos \theta +i\sin \theta )\cdots \left(ii\right)$
Putting $\left(ii\right)$ in $\left(i\right)$
$\left|2\right(r(\cos \theta +i\sin \theta ))+\frac{1}{r(\cos \theta +i\sin \theta )}|=1$
$\Rightarrow {(2r+\frac{1}{r})}^2{\cos }^2\theta +{(2r-\frac{1}{r})}^2{\sin }^2\theta =1$
$\Rightarrow {(2r+\frac{1}{r})}^2-1=({(2r+\frac{1}{r})}^2-{(2r-\frac{1}{r})}^2){\sin }^2\theta$
$\Rightarrow 4r^2+\frac{1}{r^2}+3=8{\sin }^2\theta$
$\Rightarrow 4r^2+\frac{1}{r^2}=8{\sin }^2\theta -3$
By using $A.M.\ge G.M.$, we have
$\frac{4r^2+\frac{1}{r^2}}{2}\ge \sqrt{4r^2\times \frac{1}{r^2}}$
$\Rightarrow 4r^2+\frac{1}{r^2}\ge 4$

$\therefore 8{\sin }^2\theta -3\ge 4$
$\Rightarrow 8{\sin }^2\theta \ge 7$