Solve 1/7-3i

$$\frac { 1 } { 7 - 3 i }$$

Evaluate

$$\frac{7}{58}+\frac{3}{58}i\approx 0.120689655+0.051724138i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 7+3i.

$$\frac{1\left(7+3i\right)}{\left(7-3i\right)\left(7+3i\right)}$$

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

$$\frac{1\left(7+3i\right)}{7^{2}-3^{2}i^{2}}$$

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

$$\frac{1\left(7+3i\right)}{58}$$

Multiply 1 and 7+3i to get 7+3i.

$$\frac{7+3i}{58}$$

Divide 7+3i by 58 to get \frac{7}{58}+\frac{3}{58}i.

$$\frac{7}{58}+\frac{3}{58}i$$

Real Part

$$\frac{7}{58} = 0.1206896551724138$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{7-3i} by the complex conjugate of the denominator, 7+3i.

$$Re(\frac{1\left(7+3i\right)}{\left(7-3i\right)\left(7+3i\right)})$$

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

$$Re(\frac{1\left(7+3i\right)}{7^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{1\left(7+3i\right)}{58})$$

Multiply 1 and 7+3i to get 7+3i.

$$Re(\frac{7+3i}{58})$$

Divide 7+3i by 58 to get \frac{7}{58}+\frac{3}{58}i.

$$Re(\frac{7}{58}+\frac{3}{58}i)$$

The real part of \frac{7}{58}+\frac{3}{58}i is \frac{7}{58}.

$$\frac{7}{58}$$