Solve (1+i/1-i)^3-(1-i/1+i)^3

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

Evaluate

$$-2i$$

Solution Steps

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

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

Do the multiplications in \frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

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

Divide 2i by 2 to get i.

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

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

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

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

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

Do the multiplications in \frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide -2i by 2 to get -i.

$$-i-\left(-i\right)^{3}$$

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

$$-i-i$$

Subtract i from -i to get -2i.

$$-2i$$

Real Part

$$0$$

Solution Steps

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

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

Do the multiplications in \frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

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

Divide 2i by 2 to get i.

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

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

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

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

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

Do the multiplications in \frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide -2i by 2 to get -i.

$$Re(-i-\left(-i\right)^{3})$$

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

$$Re(-i-i)$$

Subtract i from -i to get -2i.

$$Re(-2i)$$

The real part of -2i is 0.

$$0$$