Let z be a complex number such that the real part of z-2 i/z+2 i is zero. Then, the maximum value of z -(6+8 i ) is equal to :

Mathematics · Complex Numbers

JEE Main 2024 — 9 April, Shift 2 — Question 4

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 $|\mathrm{z}-(6+8 \mathrm{i})|$ is equal to :

Option A:
12
Correct
Option B:
$\infty$
Option C:
10
Option D:
8

Answer: A

Step-by-step solution

$\frac{z-2 i}{z+2 i}+\frac{\bar{z}+2 i}{\bar{z}-2 i}=0$

$\mathrm{z} \overline{\mathrm{z}}-2 \mathrm{i} \overline{\mathrm{z}}-2 \mathrm{iz}+4(-1)$

$+\mathrm{z} \overline{\mathrm{z}}+2 \mathrm{zi}+2 \overline{\mathrm{zi}}+4(-1)=0$

$\Rightarrow 2|\mathrm{z}|^{2}=8 \Rightarrow|\mathrm{z}|=2$

$|\mathrm{z}-(6+8 \mathrm{i})|_{\text {maximum }}=10+2=12$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Mathematics
Chapter: Complex Numbers
Topic: Geometry of Complex Numbers