Let S = {z in mathbb{C} - {i, 2i} : frac{z^2 | vedclass

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(A) Given the expression $f(z) = \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}$.Performing polynomial division: $f(z) = 1 + \frac{11iz - 13}{z^2

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$1680$

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(A) Given the expression $f(z) = \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}$.Performing polynomial division: $f(z) = 1 + \frac{11iz - 13}{z^2 - 3iz - 2}$.For $f(z)$ to be real,the imaginary part of the expression must be zero.Let $z = \alpha - \frac{13}{11}i$. Here $x = \alpha$ and $y = -\frac{13}{11}$.The denominator is $D = z^2 - 3iz - 2 = (x+iy)^2 - 3i(x+iy) - 2 = (x^2 - y^2 + 3y - 2) + i(2xy - 3x)$.The numerator is $N = 11iz - 13 = 11i(x+iy) - 13 = (11ix - 11y - 13) = (-11y - 13) + i(11x)$.For $\frac{N}{D} \in \mathbb{R}$,we must have $	ext{Im}(\frac{N}{D}) = 0$,which implies $	ext{Re}(N)	ext{Im}(D) = 	ext{Im}(N)	ext{Re}(D)$.Since $y = -\frac{13}{11}$,$	ext{Re}(N) = -11(-\frac{13}{11}) - 13 = 13 - 13 = 0$.Thus,for the ratio to be real,either $	ext{Im}(N) = 0$ (which implies $x = 0$,but $\alpha 
eq 0$) or $	ext{Re}(D) = 0$.Setting $	ext{Re}(D) = x^2 - y^2 + 3y - 2 = 0$:$\alpha^2 = y^2 - 3y + 2 = (y-1)(y-2)$.Substituting $y = -\frac{13}{11}$:$\alpha^2 = (-\frac{13}{11} - 1)(-\frac{13}{11} - 2) = (-\frac{24}{11})(-\frac{35}{11}) = \frac{840}{121}$.Therefore,$242\alpha^2 = 242 	imes \frac{840}{121} = 2 	imes 840 = 1680$.

Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \}$. If $\alpha - \frac{13}{11}i \in S$ and $\alpha \in \mathbb{R} - \{0\}$,then $242\alpha^2$ is equal to

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