Solve 1+1/1+2i

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

Evaluate

$$\frac{6}{5}-\frac{2}{5}i=1.2-0.4i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

Combine the real and imaginary parts in numbers 1 and \frac{1}{5}-\frac{2}{5}i.

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

Add 1 to \frac{1}{5}.

$$\frac{6}{5}-\frac{2}{5}i$$

Real Part

$$\frac{6}{5} = 1\frac{1}{5} = 1.2$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

Combine the real and imaginary parts in numbers 1 and \frac{1}{5}-\frac{2}{5}i.

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

Add 1 to \frac{1}{5}.

$$Re(\frac{6}{5}-\frac{2}{5}i)$$

The real part of \frac{6}{5}-\frac{2}{5}i is \frac{6}{5}.

$$\frac{6}{5}$$