Solve 1/4+2i

$$\frac { 1 } { 4 + 2 i }$$

Evaluate

$$\frac{1}{5}-\frac{1}{10}i=0.2-0.1i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4-2i.

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

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

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

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

$$\frac{1\left(4-2i\right)}{20}$$

Multiply 1 and 4-2i to get 4-2i.

$$\frac{4-2i}{20}$$

Divide 4-2i by 20 to get \frac{1}{5}-\frac{1}{10}i.

$$\frac{1}{5}-\frac{1}{10}i$$

Real Part

$$\frac{1}{5} = 0.2$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{4+2i} by the complex conjugate of the denominator, 4-2i.

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

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

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

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

$$Re(\frac{1\left(4-2i\right)}{20})$$

Multiply 1 and 4-2i to get 4-2i.

$$Re(\frac{4-2i}{20})$$

Divide 4-2i by 20 to get \frac{1}{5}-\frac{1}{10}i.

$$Re(\frac{1}{5}-\frac{1}{10}i)$$

The real part of \frac{1}{5}-\frac{1}{10}i is \frac{1}{5}.

$$\frac{1}{5}$$