Reduce left(frac{1}{1-4 i}-frac{2}{1+i}right) | vedclass

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Question

(A) First,simplify the expression inside the first bracket: $\frac{1}{1-4 i}-\frac{2}{1+i} = \frac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)} = \frac{1+i-2+8 i}{1+

Vedclass · Accepted Answer

$\frac{307}{442}+\frac{599 i}{442}$

Vedclass · Answer

(A) First,simplify the expression inside the first bracket: $\frac{1}{1-4 i}-\frac{2}{1+i} = \frac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)} = \frac{1+i-2+8 i}{1+i-4 i-4 i^{2}} = \frac{-1+9 i}{5-3 i}$.Next,multiply this result by the second term: $\left(\frac{-1+9 i}{5-3 i}\right)\left(\frac{3-4 i}{5+i}\right) = \frac{(-1+9 i)(3-4 i)}{(5-3 i)(5+i)}$.Expand the numerator: $(-1)(3) + (-1)(-4 i) + (9 i)(3) + (9 i)(-4 i) = -3 + 4 i + 27 i - 36 i^{2} = -3 + 31 i + 36 = 33 + 31 i$.Expand the denominator: $(5)(5) + (5)(i) + (-3 i)(5) + (-3 i)(i) = 25 + 5 i - 15 i - 3 i^{2} = 25 - 10 i + 3 = 28 - 10 i$.Now,we have $\frac{33+31 i}{28-10 i}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator,$28+10 i$: $\frac{(33+31 i)(28+10 i)}{(28-10 i)(28+10 i)} = \frac{924 + 330 i + 868 i + 310 i^{2}}{28^{2} + 10^{2}} = \frac{924 + 1198 i - 310}{784 + 100} = \frac{614 + 1198 i}{884}$.Divide both parts by $2$: $\frac{307}{442} + \frac{599}{442} i$.

Reduce $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$ to the standard form.

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