Solve 1+2i+1/2-i

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

Evaluate

$$\frac{7}{5}+\frac{11}{5}i=1.4+2.2i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

Add 1 to \frac{2}{5}. Add 2 to \frac{1}{5}.

$$\frac{7}{5}+\frac{11}{5}i$$

Real Part

$$\frac{7}{5} = 1\frac{2}{5} = 1.4$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

Add 1 to \frac{2}{5}. Add 2 to \frac{1}{5}.

$$Re(\frac{7}{5}+\frac{11}{5}i)$$

The real part of \frac{7}{5}+\frac{11}{5}i is \frac{7}{5}.

$$\frac{7}{5}$$