Solve (1-i/1+i)^13

$$( \frac { 1 - i } { 1 + i } ) ^ { 13 }$$

Evaluate

$$-i$$

Solution Steps

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

$$\left(\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}\right)^{13}$$

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

$$\left(\frac{-2i}{2}\right)^{13}$$

Divide -2i by 2 to get -i.

$$\left(-i\right)^{13}$$

Calculate -i to the power of 13 and get -i.

$$-i$$

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.

$$Re(\left(\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}\right)^{13})$$

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

$$Re(\left(\frac{-2i}{2}\right)^{13})$$

Divide -2i by 2 to get -i.

$$Re(\left(-i\right)^{13})$$

Calculate -i to the power of 13 and get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$