Question

The maximum value of |z| satisfying the equation $\frac{1}{12}(z+\overline{z}{)}^2=1-\frac{1}{3}|z{|}^2$ is:

• A.
• B.
• C. 4
• D. 6

Answer 1

The correct option is B

Let $z=r{e}^{i\theta }$.Then we have

$\frac{r^2}{12}(e^{i\theta }+e^{-i\theta }{)}^2=1-\frac{1}{3}{r}^2$

$\Rightarrow \frac{r^{2}co{s}^2\theta }{3}=1-\frac{r^2}{3}$

$\Rightarrow r^2=\frac{3}{co{s}^2\theta +1}\le 3$

$\Rightarrow max\left(r\right)=\sqrt{3}$ i.e., max |z| = $\sqrt{3}$