Let z in mathbb{C} with Im(z) = 10 and it sat | vedclass

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(C) Let $z = x + 10i$.Given $\frac{2z - n}{2z + n} = 2i - 1$.Substituting $z = x + 10i$:$\frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1$$(2x - n) + 20

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$n = 40$ and $Re(z) = -10$

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(C) Let $z = x + 10i$.Given $\frac{2z - n}{2z + n} = 2i - 1$.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 = 2i(2x + n) + 40i^2 - (2x + n) - 20i$$(2x - n) + 20i = (2i(2x + n) - 20i) - 40 - (2x + n)$Equating real and imaginary parts:Real part: $2x - n = -40 - (2x + n)$ $\Rightarrow$ $2x - n = -40 - 2x - n$ $\Rightarrow$ $4x = -40$ $\Rightarrow$ $x = -10$.Imaginary part: $20 = 2(2x + n) - 20$ $\Rightarrow$ $40 = 2(2x + n)$ $\Rightarrow$ $20 = 2x + n$.Substituting $x = -10$ into $20 = 2x + n$:$20 = 2(-10) + n$ $\Rightarrow$ $20 = -20 + n$ $\Rightarrow$ $n = 40$.Thus, $n = 40$ and $Re(z) = -10$.

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

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