The modulus of frac{1-i}{3+i}+frac{4i}{5} is | vedclass

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(C) First, simplify the expression: $\frac{1-i}{3+i} + \frac{4i}{5} = \frac{5(1-i) + 4i(3+i)}{5(3+i)}$$= \frac{5 - 5i + 12i + 4i^2}{5(3+i)}$Since $i^2

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$\frac{\sqrt{5}}{5}$ unit

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(C) First, simplify the expression: $\frac{1-i}{3+i} + \frac{4i}{5} = \frac{5(1-i) + 4i(3+i)}{5(3+i)}$$= \frac{5 - 5i + 12i + 4i^2}{5(3+i)}$Since $i^2 = -1$, we have $\frac{5 + 7i - 4}{15 + 5i} = \frac{1 + 7i}{15 + 5i}$Multiply the numerator and denominator by the conjugate $(15 - 5i)$:$= \frac{(1 + 7i)(15 - 5i)}{(15 + 5i)(15 - 5i)} = \frac{15 - 5i + 105i - 35i^2}{225 + 25} = \frac{15 + 100i + 35}{250} = \frac{50 + 100i}{250} = \frac{1 + 2i}{5}$Now, find the modulus: $|\frac{1}{5} + \frac{2}{5}i| = \sqrt{(\frac{1}{5})^2 + (\frac{2}{5})^2} = \sqrt{\frac{1}{25} + \frac{4}{25}} = \sqrt{\frac{5}{25}} = \frac{\sqrt{5}}{5}$ unit.

The modulus of $\frac{1-i}{3+i}+\frac{4i}{5}$ is

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