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
Solution
<p>$$\begin{aligned}
& n=\frac{z-2 i}{z+2 i} \\
& \text { Let } z=x+i y \\
& n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \\
& \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \\
& \Rightarrow x^2+(y-2)(y+2)=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow x^2+y^2-4=0 \\
& \Rightarrow x^2+y^2=4 \\
& \text { also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \\
& |z-(6+8 i)| \leq 2+10=12
\end{aligned}$$</p>
<p>Hence, Maximum value of $|z-(6+8 i)|$ is 12.</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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