If frac{5(-8 + 6i)}{(1 + i)^2} = a + ib,then | vedclass

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(A) Given the expression: $\frac{5(-8 + 6i)}{(1 + i)^2} = a + ib$ First,expand the denominator: $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$ Now,sub

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$(15, 20)$

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(A) Given the expression: $\frac{5(-8 + 6i)}{(1 + i)^2} = a + ib$ First,expand the denominator: $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$ Now,substitute this back into the expression: $\frac{5(-8 + 6i)}{2i} = a + ib$ Simplify the numerator: $\frac{-40 + 30i}{2i} = a + ib$ Divide each term by $2i$: $\frac{-40}{2i} + \frac{30i}{2i} = a + ib$ Since $\frac{1}{i} = -i$,we have: $-20(-i) + 15 = a + ib$ $20i + 15 = a + ib$ Rearranging gives: $15 + 20i = a + ib$ Equating real and imaginary parts,we get $a = 15$ and $b = 20$.

If $\frac{5(-8 + 6i)}{(1 + i)^2} = a + ib$,then $(a, b)$ equals

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