Solve 1/3-5i

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

Evaluate

$$\frac{3}{34}+\frac{5}{34}i\approx 0.088235294+0.147058824i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+5i.

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

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

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

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

$$\frac{1\left(3+5i\right)}{34}$$

Multiply 1 and 3+5i to get 3+5i.

$$\frac{3+5i}{34}$$

Divide 3+5i by 34 to get \frac{3}{34}+\frac{5}{34}i.

$$\frac{3}{34}+\frac{5}{34}i$$

Real Part

$$\frac{3}{34} = 0.08823529411764706$$

Solution Steps

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

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

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

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

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

$$Re(\frac{1\left(3+5i\right)}{34})$$

Multiply 1 and 3+5i to get 3+5i.

$$Re(\frac{3+5i}{34})$$

Divide 3+5i by 34 to get \frac{3}{34}+\frac{5}{34}i.

$$Re(\frac{3}{34}+\frac{5}{34}i)$$

The real part of \frac{3}{34}+\frac{5}{34}i is \frac{3}{34}.

$$\frac{3}{34}$$