Let $\mathrm{z} \in \mathrm{C}$ with $\operatorname{Im}(\mathrm{z})=10$ and it satisfies $\frac{2…

Let $\mathrm{z} \in \mathrm{C}$ with $\operatorname{Im}(\mathrm{z})=10$ and it satisfies $\frac{2 \mathrm{z}-\mathrm{n}}{2 \mathrm{z}+\mathrm{n}}=2 \mathrm{i}-1, \mathrm{i}=\sqrt{-1}$ for some natural number $\mathrm{n}$, then
  1. $\mathrm{n}=20$ and $\operatorname{Re}(\mathrm{z})=-10$
  2. $\mathrm{n}=40$ and $\operatorname{Re}(\mathrm{z})=-10$
  3. $\mathrm{n}=40$ and $\operatorname{Re}(\mathrm{z})=10$
  4. $\mathrm{n}=20$ and $\operatorname{Re}(\mathrm{z})=10$

Solution

$\begin{aligned} & \operatorname{Im}(\mathrm{z})=10 \\ & \text { Let } \mathrm{z}=x+10 \mathrm{i} \\ & \frac{2 \mathrm{z}-\mathrm{n}}{2 \mathrm{z}+\mathrm{n}}=2 \mathrm{i}-1 \\ & \Rightarrow \frac{2(x+10 \mathrm{i})-\mathrm{n}}{2(x+10 \mathrm{i})+\mathrm{n}}=2 \mathrm{i}-1 \\ & \Rightarrow(2 x-\mathrm{n})+20 \mathrm{i}=(2 \mathrm{i}-1)(2 x+20 \mathrm{i}+\mathrm{n}) \\ & \Rightarrow(2 x-\mathrm{n})+20 \mathrm{i}=(-2 x-\mathrm{n}-40)+(4 x+2 \mathrm{n}-20) \mathrm{i} \end{aligned}$ Equating real and imaginary parts, we get $\begin{aligned} & 2 x-\mathrm{n}=-2 x-\mathrm{n}-40 \text { and } 20=4 x+2 \mathrm{n}-20 \\ & \Rightarrow x=-10 \text { and } 20=4(-10)+2 \mathrm{n}-20 \\ & \Rightarrow x=-10 \text { and } \mathrm{n}=40 \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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