Question

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

• A. maximum$(|z_1+i{z}_2|)=17$
• B. minimum $(|z_1+(1+i\left)z_2|\right)=13-9\sqrt{2}$
• C. minimum$\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|=\frac{13}{4}$
• D. maximum$\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|=\frac{13}{3}$

Answer 1

Answer: (A), (D)

The correct options are
A maximum$(|z_1+i{z}_2|)=17$
D maximum$\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|=\frac{13}{3}$

$z_1=5+12i,|z_2|=4$
$|z_1+i{z}_2|\le |z_1|+|z_2|=13+4=17$
$\therefore |z_1+(1+i)z_2|\ge ||z_1|-|1+i||z_2||$
$=13-4\sqrt{2}$
$\therefore min(|z_1+(1+i\left)z_2|\right)=13-4\sqrt{2}$
$|z_2+\frac{4}{z_2}|\le |z_2|+\frac{4}{|z_2|}=4+1=5$
$|z_2+\frac{4}{z_2}|\ge |z_2|-\frac{4}{|z_2|}=4-1=3$
$\therefore max\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|=\frac{13}{3}andmin\left|\frac{z_1}{z_2+\frac{4}{z_2}}\right|=\frac{13}{5}$