Solve 1-i/1-3i+2-i/3+i

$$\frac { 1 - i } { 1 - 3 i } + \frac { 2 - i } { 3 + i }$$

Evaluate

$$\frac{9}{10}-\frac{3}{10}i=0.9-0.3i$$

Solution Steps

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

$$\frac{\left(1-i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}+\frac{2-i}{3+i}$$

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

$$\frac{4+2i}{10}+\frac{2-i}{3+i}$$

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

$$\frac{2}{5}+\frac{1}{5}i+\frac{2-i}{3+i}$$

Multiply both numerator and denominator of \frac{2-i}{3+i} by the complex conjugate of the denominator, 3-i.

$$\frac{2}{5}+\frac{1}{5}i+\frac{\left(2-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$$

Do the multiplications in \frac{\left(2-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}.

$$\frac{2}{5}+\frac{1}{5}i+\frac{5-5i}{10}$$

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

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

Add \frac{2}{5}+\frac{1}{5}i and \frac{1}{2}-\frac{1}{2}i to get \frac{9}{10}-\frac{3}{10}i.

$$\frac{9}{10}-\frac{3}{10}i$$

Real Part

$$\frac{9}{10} = 0.9$$

Solution Steps

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

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

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

$$Re(\frac{4+2i}{10}+\frac{2-i}{3+i})$$

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

$$Re(\frac{2}{5}+\frac{1}{5}i+\frac{2-i}{3+i})$$

Multiply both numerator and denominator of \frac{2-i}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{2}{5}+\frac{1}{5}i+\frac{\left(2-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)})$$

Do the multiplications in \frac{\left(2-i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}.

$$Re(\frac{2}{5}+\frac{1}{5}i+\frac{5-5i}{10})$$

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

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

Add \frac{2}{5}+\frac{1}{5}i and \frac{1}{2}-\frac{1}{2}i to get \frac{9}{10}-\frac{3}{10}i.

$$Re(\frac{9}{10}-\frac{3}{10}i)$$

The real part of \frac{9}{10}-\frac{3}{10}i is \frac{9}{10}.

$$\frac{9}{10}$$