Question

If $|z-\frac{4}{2}|=2,$ then the maximum value of $|Z|$ is equal to

• A. $2$
• B. $2+\sqrt{2}$
• C. $\sqrt{3}+1$
• D. $\sqrt{5}+1$

Answer 1

The correct option is D $\sqrt{5}+1$

$\begin{array}{c}Letz\in c\\ wehave\\ z=\rho e^{i\varphi }\\ weknowthat\\ \left|z\right|=\sqrt{z\overline{z}}\\ \left(\rho e^{i\varphi }-\frac{4}{\rho }{e}^{-i\varphi }\right)-\left(\rho e^{i\varphi }-\frac{4}{\rho }{e}^{i\varphi }\right)=4\\ p^2-8\cos \left(2\varphi \right)+\frac{16}{{\rho }^2}=4\\ {\rho }^4-2\left(2+4\cos \left(2\varphi \right)\right){\rho }^2+16=0\\ Now,solvingfor{\rho }^2\\ {\rho }^2=2\left(2+4\cos \left(2\varphi \right)\pm \sqrt{4\cos \left(2\varphi \right)+2\cos \left(4\varphi \right)-1}\right)\\ For\varphi =0\\ {{\rho }^2}_{max}=2\left(1+2+\sqrt{4+2-1}\right)=2\left(3+\sqrt{5}\right)\\ then\\ maz\left|z\right|=\sqrt{2\left(3+\sqrt{5}\right)}=1+\sqrt{5}\\ Hence,optionDistherequiredanswer.\end{array}$