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

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