text{Re} left( frac{(1 + i)^2}{3 - i} right) | vedclass

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Question

(A) First,simplify the numerator: $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$. Now,the expression becomes $\frac{2i}{3 - i}$. To simplify this,mult

Vedclass · Accepted Answer

$-1/5$

Vedclass · Answer

(A) First,simplify the numerator: $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$. Now,the expression becomes $\frac{2i}{3 - i}$. To simplify this,multiply the numerator and denominator by the conjugate of the denominator,which is $(3 + i)$: $\frac{2i(3 + i)}{(3 - i)(3 + i)} = \frac{6i + 2i^2}{3^2 - i^2} = \frac{6i - 2}{9 + 1} = \frac{-2 + 6i}{10} = -\frac{2}{10} + \frac{6}{10}i = -\frac{1}{5} + \frac{3}{5}i$. The real part $	ext{Re}$ of this complex number is $-\frac{1}{5}$.

$\text{Re} \left( \frac{(1 + i)^2}{3 - i} \right) =$

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