Let z in mathbb{C} with operatorname{Im}(z)=1 | vedclass

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(B) Given $\operatorname{Im}(z)=10$, let $z=x+10i$.The given equation is $\frac{2z-n}{2z+n}=2i-1$.Substituting $z=x+10i$:$\frac{2(x+10i)-n}{2(x+10i)+n

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$n=40$ and $\operatorname{Re}(z)=-10$

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(B) Given $\operatorname{Im}(z)=10$, let $z=x+10i$.The given equation is $\frac{2z-n}{2z+n}=2i-1$.Substituting $z=x+10i$:$\frac{2(x+10i)-n}{2(x+10i)+n}=2i-1$$(2x-n)+20i=(2i-1)(2x+n+20i)$$(2x-n)+20i = 4xi + 2ni - 40 - 2x - n - 20i$$(2x-n)+20i = (-2x-n-40) + (4x+2n-20)i$Equating real and imaginary parts:Real part: $2x-n = -2x-n-40$ $\Rightarrow 4x = -40$ $\Rightarrow x = -10$.Imaginary part: $20 = 4x+2n-20$.Substituting $x=-10$: $20 = 4(-10)+2n-20$ $\Rightarrow 20 = -40+2n-20$ $\Rightarrow 2n = 80$ $\Rightarrow n = 40$.Thus, $n=40$ and $\operatorname{Re}(z)=x=-10$.

Let $z \in \mathbb{C}$ with $\operatorname{Im}(z)=10$ and it satisfies $\frac{2z-n}{2z+n}=2i-1$, where $i=\sqrt{-1}$, for some natural number $n$. Then:

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