Solve 5/1-3i

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

Evaluate

$$\frac{1}{2}+\frac{3}{2}i=0.5+1.5i$$

Solution Steps

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

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

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

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

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

$$\frac{5\left(1+3i\right)}{10}$$

Multiply 5 times 1+3i.

$$\frac{5\times 1+5\times \left(3i\right)}{10}$$

Do the multiplications in 5\times 1+5\times \left(3i\right).

$$\frac{5+15i}{10}$$

Divide 5+15i by 10 to get \frac{1}{2}+\frac{3}{2}i.

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

Real Part

$$\frac{1}{2} = 0.5$$

Solution Steps

Multiply both numerator and denominator of \frac{5}{1-3i} by the complex conjugate of the denominator, 1+3i.

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

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

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

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

$$Re(\frac{5\left(1+3i\right)}{10})$$

Multiply 5 times 1+3i.

$$Re(\frac{5\times 1+5\times \left(3i\right)}{10})$$

Do the multiplications in 5\times 1+5\times \left(3i\right).

$$Re(\frac{5+15i}{10})$$

Divide 5+15i by 10 to get \frac{1}{2}+\frac{3}{2}i.

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

The real part of \frac{1}{2}+\frac{3}{2}i is \frac{1}{2}.

$$\frac{1}{2}$$