Solve 1/4+3i

$$\frac { 1 } { 4 + 3 i }$$

Evaluate

$$\frac{4}{25}-\frac{3}{25}i=0.16-0.12i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4-3i.

$$\frac{1\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$$

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

$$\frac{1\left(4-3i\right)}{4^{2}-3^{2}i^{2}}$$

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

$$\frac{1\left(4-3i\right)}{25}$$

Multiply 1 and 4-3i to get 4-3i.

$$\frac{4-3i}{25}$$

Divide 4-3i by 25 to get \frac{4}{25}-\frac{3}{25}i.

$$\frac{4}{25}-\frac{3}{25}i$$

Real Part

$$\frac{4}{25} = 0.16$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$Re(\frac{1\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)})$$

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

$$Re(\frac{1\left(4-3i\right)}{4^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{1\left(4-3i\right)}{25})$$

Multiply 1 and 4-3i to get 4-3i.

$$Re(\frac{4-3i}{25})$$

Divide 4-3i by 25 to get \frac{4}{25}-\frac{3}{25}i.

$$Re(\frac{4}{25}-\frac{3}{25}i)$$

The real part of \frac{4}{25}-\frac{3}{25}i is \frac{4}{25}.

$$\frac{4}{25}$$