JEE Mains · Maths · STD 11 - 4.1 complex nubers
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 :
- A \(12\)
- B \(\infty\)
- C \(10\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(12\)
Step-by-step Solution
Detailed explanation
\( \frac{z-2 i}{z+2 i}+\frac{\bar{z}+2 i}{\bar{z}-2 i}=0 \) \( z \bar{z}-2 i \bar{z}-2 i z+4(-1) \) \( +z \bar{z}+2 z i+2 \bar{z} i+4(-1)=0 \) \( \Rightarrow 2|z|^2=8 \Rightarrow|z|=2 \) \( |z-(6+8 i)|_{\text {maximum }}=10+2=12\)
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