Solve 2i/(1-i)^3=

$$\frac { 2 i } { ( 1 - i ) ^ { 3 } } =$$

Evaluate

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

Solution Steps

Calculate 1-i to the power of 3 and get -2-2i.

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

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

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

Do the multiplications in \frac{2i\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}.

$$\frac{-4-4i}{8}$$

Divide -4-4i by 8 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

Calculate 1-i to the power of 3 and get -2-2i.

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

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

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

Do the multiplications in \frac{2i\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}.

$$Re(\frac{-4-4i}{8})$$

Divide -4-4i by 8 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}.

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