Solve 1/(13.5+i21)

$$\frac { 1 } { ( 13.5 + i 21 ) }$$

Evaluate

$$\frac{6}{277}-\frac{28}{831}i\approx 0.02166065-0.033694344i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 13.5-21i.

$$\frac{1\left(13.5-21i\right)}{\left(13.5+21i\right)\left(13.5-21i\right)}$$

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

$$\frac{1\left(13.5-21i\right)}{13.5^{2}-21^{2}i^{2}}$$

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

$$\frac{1\left(13.5-21i\right)}{623.25}$$

Multiply 1 and 13.5-21i to get 13.5-21i.

$$\frac{13.5-21i}{623.25}$$

Divide 13.5-21i by 623.25 to get \frac{6}{277}-\frac{28}{831}i.

$$\frac{6}{277}-\frac{28}{831}i$$

Real Part

$$\frac{6}{277} = 0.021660649819494584$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{13.5+21i} by the complex conjugate of the denominator, 13.5-21i.

$$Re(\frac{1\left(13.5-21i\right)}{\left(13.5+21i\right)\left(13.5-21i\right)})$$

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

$$Re(\frac{1\left(13.5-21i\right)}{13.5^{2}-21^{2}i^{2}})$$

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

$$Re(\frac{1\left(13.5-21i\right)}{623.25})$$

Multiply 1 and 13.5-21i to get 13.5-21i.

$$Re(\frac{13.5-21i}{623.25})$$

Divide 13.5-21i by 623.25 to get \frac{6}{277}-\frac{28}{831}i.

$$Re(\frac{6}{277}-\frac{28}{831}i)$$

The real part of \frac{6}{277}-\frac{28}{831}i is \frac{6}{277}.

$$\frac{6}{277}$$