Discussion
Let z=\(\dfrac{2 - i}{\alpha + i}\) , where \(\alpha\) is a real number. If \(4Re(z) = 3Im(\bar{z})\) then the value of \(\alpha\) is
Answer: D
\(\alpha = \) 2.
Multiply numerator and denominator by the conjugate \(\alpha - i\):
\[ \begin{aligned} z &= \frac{(2 - i)(\alpha - i)}{\alpha^{2} + 1} \\ &= \frac{(2\alpha - 1) - (2 + \alpha)i}{\alpha^{2} + 1} \end{aligned} \]
So \(Re(z) = \frac{2\alpha - 1}{\alpha^{2} + 1}\) and \(Im(\bar{z}) = -Im(z) = \frac{2 + \alpha}{\alpha^{2} + 1}\). The condition gives
\[ \begin{aligned} 4(2\alpha - 1) &= 3(2 + \alpha) \\ 8\alpha - 4 &= 6 + 3\alpha \\ \alpha &= 2 \end{aligned} \]
Using \(Im(z)\) instead of \(Im(\bar{z})\) flips the sign of the right side and leads to a non-integer value, so none of -2, 3, 5 or -5 satisfies the stated condition.
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