Solve i/4-2i

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

Evaluate

$$-\frac{1}{10}+\frac{1}{5}i=-0.1+0.2i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4+2i.

$$\frac{i\left(4+2i\right)}{\left(4-2i\right)\left(4+2i\right)}$$

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

$$\frac{i\left(4+2i\right)}{4^{2}-2^{2}i^{2}}$$

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

$$\frac{i\left(4+2i\right)}{20}$$

Multiply i times 4+2i.

$$\frac{4i+2i^{2}}{20}$$

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

$$\frac{4i+2\left(-1\right)}{20}$$

Do the multiplications in 4i+2\left(-1\right). Reorder the terms.

$$\frac{-2+4i}{20}$$

Divide -2+4i by 20 to get -\frac{1}{10}+\frac{1}{5}i.

$$-\frac{1}{10}+\frac{1}{5}i$$

Real Part

$$-\frac{1}{10} = -0.1$$

Solution Steps

Multiply both numerator and denominator of \frac{i}{4-2i} by the complex conjugate of the denominator, 4+2i.

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

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

$$Re(\frac{i\left(4+2i\right)}{4^{2}-2^{2}i^{2}})$$

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

$$Re(\frac{i\left(4+2i\right)}{20})$$

Multiply i times 4+2i.

$$Re(\frac{4i+2i^{2}}{20})$$

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

$$Re(\frac{4i+2\left(-1\right)}{20})$$

Do the multiplications in 4i+2\left(-1\right). Reorder the terms.

$$Re(\frac{-2+4i}{20})$$

Divide -2+4i by 20 to get -\frac{1}{10}+\frac{1}{5}i.

$$Re(-\frac{1}{10}+\frac{1}{5}i)$$

The real part of -\frac{1}{10}+\frac{1}{5}i is -\frac{1}{10}.

$$-\frac{1}{10}$$