Let z be a complex number such that the real part of [(z-2 i)/(z+2 i)] is zero.

Question

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

Accepted Answer

$$\begin{aligned}
& n=\frac{z-2 i}{z+2 i} \
& \text { Let } z=x+i y \
& n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \
& \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \
& \Rightarrow x^2+(y-2)(y+2)=0
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow x^2+y^2-4=0 \
& \Rightarrow x^2+y^2=4 \
& \text { also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \
& |z-(6+8 i)| \leq 2+10=12
\end{aligned}$$

Hence, Maximum value of $|z-(6+8 i)|$ is 12.

Solutions · 1