Let $z$ be a complex number such that the real part of $\frac{z-2 i}{z+2 i}$ is zero. Then, the maximum…
Let $z$ be a complex number such that the real part of $\frac{z-2 i}{z+2 i}$ is zero. Then, the maximum value of $|z-(6+8 i)|$ is equal to
- 12
- 10
- 8
- $\infty$
Solution
$\begin{aligned} & \frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}+\frac{\overline{\mathrm{z}}+2 \mathrm{i}}{\overline{\mathrm{z}}-2 \mathrm{i}}=0 \\ & \mathrm{z} \overline{\mathrm{z}}-2 \mathrm{i} \overline{\mathrm{z}}-2 \mathrm{i} \mathrm{z}+4(-1) \\ & +\mathrm{z} \overline{\mathrm{z}}+2 \mathrm{z} i+2 \overline{\mathrm{z} i}+4(-1)=0 \\ & \Rightarrow 2|\mathrm{z}|^2=8 \Rightarrow|\mathrm{z}|=2 \\ & |\mathrm{z}-(6+8 \mathrm{i})|_{\text {maximum }}=10+2=12\end{aligned}$
Asked in: JEE Main 2024 (09 Apr Shift 2)
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