Solve 2i/1-2i

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

Evaluate

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

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+2i.

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

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

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

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

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

Multiply 2i times 1+2i.

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

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

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

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

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

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

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

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

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

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

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

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

Multiply 2i times 1+2i.

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

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

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

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

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

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}.

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