Solve i/1-i

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

Evaluate

$$-\frac{1}{2}+\frac{1}{2}i=-0.5+0.5i$$

Solution Steps

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

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

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

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

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

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

Multiply i times 1+i.

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

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

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

Reorder the terms.

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

Divide -1+i by 2 to get -\frac{1}{2}+\frac{1}{2}i.

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

Real Part

$$-\frac{1}{2} = -0.5$$

Solution Steps

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

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

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

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

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

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

Multiply i times 1+i.

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

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

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

Reorder the terms.

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

Divide -1+i by 2 to get -\frac{1}{2}+\frac{1}{2}i.

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

The real part of -\frac{1}{2}+\frac{1}{2}i is -\frac{1}{2}.

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