Solve i/2+i

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

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, 2-i.

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

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

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

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

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

Multiply i times 2-i.

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

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

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

Do the multiplications in 2i-\left(-1\right). Reorder the terms.

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

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{i}{2+i} by the complex conjugate of the denominator, 2-i.

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

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

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

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

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

Multiply i times 2-i.

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

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

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

Do the multiplications in 2i-\left(-1\right). Reorder the terms.

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

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}.

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