Solve {left(1+i/1-iright)}^99

$${ \left( \frac{ 1+i }{ 1-i } \right) }^{ 99 }$$

Evaluate

$$-i$$

Solution Steps

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

$$\left(\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{99}$$

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

$$\left(\frac{2i}{2}\right)^{99}$$

Divide 2i by 2 to get i.

$$i^{99}$$

Calculate i to the power of 99 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.

$$Re(\left(\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{99})$$

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

$$Re(\left(\frac{2i}{2}\right)^{99})$$

Divide 2i by 2 to get i.

$$Re(i^{99})$$

Calculate i to the power of 99 and get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$