The imaginary part of frac{(1 + i)^2}{2 - i} | vedclass

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(C) We have $z = \frac{(1 + i)^2}{2 - i}$.Since $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$, the expression becomes $z = \frac{2i}{2 - i}$.To simpl

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$\frac{4}{5}$

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(C) We have $z = \frac{(1 + i)^2}{2 - i}$.Since $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$, the expression becomes $z = \frac{2i}{2 - i}$.To simplify, multiply the numerator and denominator by the conjugate of the denominator $(2 + i)$:$z = \frac{2i(2 + i)}{(2 - i)(2 + i)} = \frac{4i + 2i^2}{2^2 - i^2} = \frac{4i - 2}{4 + 1} = \frac{-2 + 4i}{5} = -\frac{2}{5} + i\frac{4}{5}$.Thus, the imaginary part $Im(z) = \frac{4}{5}$.

The imaginary part of $\frac{(1 + i)^2}{2 - i}$ is

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