Solve 1/3+4i

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

Evaluate

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

Solution Steps

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

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

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

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

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

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

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

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

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

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

Real Part

$$\frac{3}{25} = 0.12$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$\frac{3}{25}$$