Let z be a complex number with operatorname{I | vedclass

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(D) Given $\frac{2z-n}{2z+n} = 2i-1$. Let $2z = x+iy$. Since $\operatorname{Im}(z)=10$, we have $\operatorname{Im}(2z) = 20$. So $2z = x+20i$. Substit

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

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(D) Given $\frac{2z-n}{2z+n} = 2i-1$. Let $2z = x+iy$. Since $\operatorname{Im}(z)=10$, we have $\operatorname{Im}(2z) = 20$. So $2z = x+20i$. Substituting this into the equation: $\frac{x+20i-n}{x+20i+n} = 2i-1$. $(x-n)+20i = (2i-1)(x+n+20i)$. $(x-n)+20i = 2i(x+n) + 40i^2 - (x+n) - 20i$. $(x-n)+20i = 2i(x+n) - 40 - (x+n) - 20i$. $(x-n)+20i = -(x+n+40) + i(2x+2n-20)$. Equating real and imaginary parts: Real part: $x-n = -(x+n+40) \implies x-n = -x-n-40 \implies 2x = -40 \implies x = -20$. Since $2z = x+20i$, we have $2z = -20+20i$, so $z = -10+10i$. Thus $\operatorname{Re}(z) = -10$. Imaginary part: $20 = 2x+2n-20$. Substituting $x=-20$: $20 = 2(-20)+2n-20 \implies 20 = -40+2n-20 \implies 20 = 2n-60 \implies 2n = 80 \implies n = 40$. Therefore, $n=40$ and $\operatorname{Re}(z)=-10$.

Let $z$ be a complex number with $\operatorname{Im}(z)=10$ and satisfying $\frac{2z-n}{2z+n}=2i-1$, where $i=\sqrt{-1}$, for some natural number $n$. Then:

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