Question 4, Review Exercise

Solutions of Question 4 of Review Exercise of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Locate the complex number $z=x+i y$ on the complex plane if $\left|\dfrac{z+2 i}{z-2 i}\right|=1$

Solution.

Given $z = x + iy$, then $$\begin{align*} & \left|\dfrac{z + 2i}{z - 2i}\right| = 1\\ \implies & |z + 2i| = |z - 2i|\\ \implies & |x + i(y + 2)| = |x + i(y - 2)|\\ \implies & \sqrt{x^2 + (y + 2)^2} = \sqrt{x^2 + (y - 2)^2} \end{align*}$$ Squaring both sides, we have $$\begin{align*} & x^2 + (y + 2)^2 = x^2 + (y - 2)^2\\ \implies & (y + 2)^2 = (y - 2)^2\\ \implies & y^2 + 4y + 4 = y^2 - 4y + 4 \\ \implies & 4y +4y =0 \\ \implies & 8y=0 \implies y=0. \end{align*}$$ Hence, we conclude $z=x+i \cdot 0$ $\implies z=x$.

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