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