Jee Mains
Let z be the complex number satisfying $|z-5| \leq 3$ and having maximum positive principal argument. Then $34\left|\frac{5 z-12}{5 i z+16}\right|^{2}$ is equal to
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Solution
$|Z-5| \leq 3$ center $C(5,0), R=3$ Point $A \equiv$ greatest argument $\Delta O A C$ is right angled triangled $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$
\[ \begin{aligned} & \mathrm{OM}=4 \cos \theta=\frac{16}{5} ; \mathrm{AM}=4 \sin \theta=\frac{12}{5} \\ & \mathrm{~A}\left(\frac{16}{5}, \frac{12}{5}\right) \mathrm{Z}=\frac{16+12 \mathrm{i}}{5} \\ & 34\left|\frac{5 \mathrm{z}-12}{5 \mathrm{zi}+16}\right|^{2}=34\left|\frac{16+12 \mathrm{i}-12}{(16+12 \mathrm{i}) \mathrm{i}+16}\right|^{2}=34\left|\frac{4+12 \mathrm{i}}{4+16 \mathrm{i}}\right|^{2}=34\left|\frac{1+3 \mathrm{i}}{1+4 \mathrm{i}}\right|^{2}=34 \frac{(1+9)}{(1+16)}=2(10)=20 \end{aligned} \]
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