Solve 1/1-2i

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

Evaluate

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

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+2i.

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

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

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

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

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

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

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

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

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

Real Part

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

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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