Jee Mains
Let z be a complex number such that the real part of $\frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}$ is zero. Then, the maximum value of $|\mathrm{z}-(6+8 \mathrm{i})|$ is equal to
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Solution
According to the question, $\operatorname{Re}\left(\frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}\right)=0 \Rightarrow \frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}+\overline{\left(\frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}\right)}=0$
\[ \begin{aligned} & \Rightarrow \frac{z-2 i}{z+2 i}+\frac{\bar{z}+2 i}{\bar{z}-2 i}=0 \Rightarrow(z-2 i)(\bar{z}-2 i)+(\bar{z}+2 i)(z+2 i)=0 \\ & \Rightarrow z \bar{z}-2 i \bar{z}-2 i z+4(-1)+z \bar{z}+2 z i+2 \bar{z} i+4(-1)=0 \\ & \Rightarrow 2|z|^{2}=8 \Rightarrow|z|=2 \Rightarrow|z-(6+8 i)|_{\text {maximum }}=10+2=12 \end{aligned} \]
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