Solve 2/1-3i

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

Evaluate

$$\frac{1}{5}+\frac{3}{5}i=0.2+0.6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+3i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{2\left(1+3i\right)}{10}$$

Multiply 2 times 1+3i.

$$\frac{2\times 1+2\times \left(3i\right)}{10}$$

Do the multiplications in 2\times 1+2\times \left(3i\right).

$$\frac{2+6i}{10}$$

Divide 2+6i by 10 to get \frac{1}{5}+\frac{3}{5}i.

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

Real Part

$$\frac{1}{5} = 0.2$$

Solution Steps

Multiply both numerator and denominator of \frac{2}{1-3i} by the complex conjugate of the denominator, 1+3i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{2\left(1+3i\right)}{10})$$

Multiply 2 times 1+3i.

$$Re(\frac{2\times 1+2\times \left(3i\right)}{10})$$

Do the multiplications in 2\times 1+2\times \left(3i\right).

$$Re(\frac{2+6i}{10})$$

Divide 2+6i by 10 to get \frac{1}{5}+\frac{3}{5}i.

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

The real part of \frac{1}{5}+\frac{3}{5}i is \frac{1}{5}.

$$\frac{1}{5}$$