Question

If $|z-\frac{4}{z}]=2$, Then the maximum value of $|z|$ is equal tp

• A. $1+\sqrt{3}$
• B. $1+\sqrt{5}$
• C. $1-\sqrt{5}$
• D. $\sqrt{5}-1$

Answer 1

Answer: (B)

$1+\sqrt{5}$

Consider the following question.

$Z=\rho e^{i\varphi }$

We know that.

$|Z|=\sqrt{Z\overline{Z}}$

$(\rho {e^i}^{\varphi }-4\rho e^{-i\varphi })(\rho e^{-i\varphi }-4\rho e^{i\varphi })=4$

${\rho }^2-8cos\left(2\varphi \right)+\frac{16}{{\rho }^2}=4$

${\rho }^4-2(2+4cos(2\varphi \left)\right){\rho }^2+16=0$

Now solving for ${\rho }^2$

${\rho }^2=2(1+2cos(2\varphi )\pm \sqrt{4cos\left(2\varphi \right)+2cos\left(4\varphi \right)-1})$

${\rho }^{2}max=2(1+2+\sqrt{4+2-1})=2(3+\sqrt{5})$

$max|Z|=\sqrt{2(3+\sqrt{5})}=1+\sqrt{5}$

Hence, this is the required answer.