Solve sqrt[3]{1-i/1+i}

$$\sqrt[ 3 ] { \frac { 1 - i } { 1 + i } }$$

Evaluate

$$i$$

Solution Steps

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

$$\sqrt[3]{\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}}$$

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

$$\sqrt[3]{\frac{-2i}{2}}$$

Divide -2i by 2 to get -i.

$$\sqrt[3]{-i}$$

Calculate \sqrt[3]{-i} 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.

$$Re(\sqrt[3]{\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}})$$

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

$$Re(\sqrt[3]{\frac{-2i}{2}})$$

Divide -2i by 2 to get -i.

$$Re(\sqrt[3]{-i})$$

Calculate \sqrt[3]{-i} and get i.

$$Re(i)$$

The real part of i is 0.

$$0$$