Solve 1+1/2-i

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

Evaluate

$$\frac{4}{5}+\frac{2}{5}i=0.8+0.4i$$

Solution Steps

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

Multiply 2 times 2+i.

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

Do the multiplications in 2\times 2+2i.

$$\frac{4+2i}{5}$$

Divide 4+2i by 5 to get \frac{4}{5}+\frac{2}{5}i.

$$\frac{4}{5}+\frac{2}{5}i$$

Real Part

$$\frac{4}{5} = 0.8$$

Solution Steps

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

Multiply 2 times 2+i.

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

Do the multiplications in 2\times 2+2i.

$$Re(\frac{4+2i}{5})$$

Divide 4+2i by 5 to get \frac{4}{5}+\frac{2}{5}i.

$$Re(\frac{4}{5}+\frac{2}{5}i)$$

The real part of \frac{4}{5}+\frac{2}{5}i is \frac{4}{5}.

$$\frac{4}{5}$$