Solve 1-i/3+i*4i/5

$$\frac { 1 - i } { 3 + i } \cdot \frac { 4 i } { 5 }$$

Evaluate

$$\frac{8}{25}+\frac{4}{25}i=0.32+0.16i$$

Solution Steps

Multiply both numerator and denominator of \frac{1-i}{3+i} by the complex conjugate of the denominator, 3-i.

$$\frac{\left(1-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}\times \frac{4i}{5}$$

Do the multiplications in \frac{\left(1-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}.

$$\frac{2-4i}{10}\times \frac{4i}{5}$$

Divide 2-4i by 10 to get \frac{1}{5}-\frac{2}{5}i.

$$\left(\frac{1}{5}-\frac{2}{5}i\right)\times \frac{4i}{5}$$

Divide 4i by 5 to get \frac{4}{5}i.

$$\left(\frac{1}{5}-\frac{2}{5}i\right)\times \left(\frac{4}{5}i\right)$$

Multiply \frac{1}{5}-\frac{2}{5}i and \frac{4}{5}i to get \frac{8}{25}+\frac{4}{25}i.

$$\frac{8}{25}+\frac{4}{25}i$$

Real Part

$$\frac{8}{25} = 0.32$$

Solution Steps

Multiply both numerator and denominator of \frac{1-i}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{\left(1-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}\times \frac{4i}{5})$$

Do the multiplications in \frac{\left(1-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}.

$$Re(\frac{2-4i}{10}\times \frac{4i}{5})$$

Divide 2-4i by 10 to get \frac{1}{5}-\frac{2}{5}i.

$$Re(\left(\frac{1}{5}-\frac{2}{5}i\right)\times \frac{4i}{5})$$

Divide 4i by 5 to get \frac{4}{5}i.

$$Re(\left(\frac{1}{5}-\frac{2}{5}i\right)\times \left(\frac{4}{5}i\right))$$

Multiply \frac{1}{5}-\frac{2}{5}i and \frac{4}{5}i to get \frac{8}{25}+\frac{4}{25}i.

$$Re(\frac{8}{25}+\frac{4}{25}i)$$

The real part of \frac{8}{25}+\frac{4}{25}i is \frac{8}{25}.

$$\frac{8}{25}$$