Solve (1/1-i)^8

$$( \frac { 1 } { 1 - i } ) ^ { 8 }$$

Evaluate

$$\frac{1}{16}=0.0625$$

Solution Steps

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

$$\left(\frac{1\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{8}$$

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

$$\left(\frac{1+i}{2}\right)^{8}$$

Divide 1+i by 2 to get \frac{1}{2}+\frac{1}{2}i.

$$\left(\frac{1}{2}+\frac{1}{2}i\right)^{8}$$

Calculate \frac{1}{2}+\frac{1}{2}i to the power of 8 and get \frac{1}{16}.

$$\frac{1}{16}$$

Real Part

$$\frac{1}{16} = 0.0625$$

Solution Steps

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

$$Re(\left(\frac{1\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{8})$$

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

$$Re(\left(\frac{1+i}{2}\right)^{8})$$

Divide 1+i by 2 to get \frac{1}{2}+\frac{1}{2}i.

$$Re(\left(\frac{1}{2}+\frac{1}{2}i\right)^{8})$$

Calculate \frac{1}{2}+\frac{1}{2}i to the power of 8 and get \frac{1}{16}.

$$Re(\frac{1}{16})$$

The real part of \frac{1}{16} is \frac{1}{16}.

$$\frac{1}{16}$$