Solve 3i/1-i

$$\frac { 3 i } { 1 - i }$$

Evaluate

$$-\frac{3}{2}+\frac{3}{2}i=-1.5+1.5i$$

Solution Steps

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

$$\frac{3i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}$$

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

$$\frac{3i\left(1+i\right)}{1^{2}-i^{2}}$$

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

$$\frac{3i\left(1+i\right)}{2}$$

Multiply 3i times 1+i.

$$\frac{3i\times 1+3i^{2}}{2}$$

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

$$\frac{3i\times 1+3\left(-1\right)}{2}$$

Do the multiplications in 3i\times 1+3\left(-1\right). Reorder the terms.

$$\frac{-3+3i}{2}$$

Divide -3+3i by 2 to get -\frac{3}{2}+\frac{3}{2}i.

$$-\frac{3}{2}+\frac{3}{2}i$$

Real Part

$$-\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

Multiply both numerator and denominator of \frac{3i}{1-i} by the complex conjugate of the denominator, 1+i.

$$Re(\frac{3i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)})$$

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

$$Re(\frac{3i\left(1+i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{3i\left(1+i\right)}{2})$$

Multiply 3i times 1+i.

$$Re(\frac{3i\times 1+3i^{2}}{2})$$

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

$$Re(\frac{3i\times 1+3\left(-1\right)}{2})$$

Do the multiplications in 3i\times 1+3\left(-1\right). Reorder the terms.

$$Re(\frac{-3+3i}{2})$$

Divide -3+3i by 2 to get -\frac{3}{2}+\frac{3}{2}i.

$$Re(-\frac{3}{2}+\frac{3}{2}i)$$

The real part of -\frac{3}{2}+\frac{3}{2}i is -\frac{3}{2}.

$$-\frac{3}{2}$$