Solve 2i/2-i

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

Evaluate

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

Solution Steps

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

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

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

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

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

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

Multiply 2i times 2+i.

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

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

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

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

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

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

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

Real Part

$$-\frac{2}{5} = -0.4$$

Solution Steps

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

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

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

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

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

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

Multiply 2i times 2+i.

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

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

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

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

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

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

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

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

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