Solve 1/2-i

$$\frac { 1 } { 2 - i }$$

Evaluate

$$\frac{2}{5}+\frac{1}{5}i=0.4+0.2i$$

Solution Steps

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

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

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

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

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

$$\frac{1\left(2+i\right)}{5}$$

Multiply 1 and 2+i to get 2+i.

$$\frac{2+i}{5}$$

Divide 2+i by 5 to get \frac{2}{5}+\frac{1}{5}i.

$$\frac{2}{5}+\frac{1}{5}i$$

Real Part

$$\frac{2}{5} = 0.4$$

Solution Steps

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

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

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

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

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

$$Re(\frac{1\left(2+i\right)}{5})$$

Multiply 1 and 2+i to get 2+i.

$$Re(\frac{2+i}{5})$$

Divide 2+i by 5 to get \frac{2}{5}+\frac{1}{5}i.

$$Re(\frac{2}{5}+\frac{1}{5}i)$$

The real part of \frac{2}{5}+\frac{1}{5}i is \frac{2}{5}.

$$\frac{2}{5}$$