Question

If $z_1=5+12i$ and $|z_2|=4$, then

• A. Maximum value of $|z_1+i{z}_2|$ is $17$
• B. Minimum value of $|z_1+(1+i)z_2|$ is $13-4\sqrt{2}$
• C. Minimum value of $\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|$ is $\frac{13}{4}$
• D. Maximum value of $\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|$ is $\frac{13}{3}$

Answer 1

Answer: (A), (B), (D)

Maximum value of $\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|$ is $\frac{13}{3}$

Here,
$|z_1+i{z}_2|\le |z_1|+|i{z}_2|\le 13+4\le 17$

$|z_1+(1+i)z_2|\ge ||z_1|-|(1+i)z_2||\ge 13-4\sqrt{2}$

Now,
$||z_2|-\frac{4}{|z_2|}|\le |z_2+\frac{4}{z_2}|\le |z_2|+\frac{4}{|z_2|}\Rightarrow 3\le |z_2+\frac{4}{z_2}|\le 5\therefore \left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|\in (\frac{13}{5},\frac{13}{3})(\because |z_1|=13)$
So, the minimum value is $\frac{13}{5}$ and the maximum value is $\frac{13}{3}$