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

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

Evaluate

$$11+2i$$

Solution Steps

Multiply both numerator and denominator of \frac{1-3i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(1-3i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$\left(\frac{-2-4i}{2}\right)^{3}$$

Divide -2-4i by 2 to get -1-2i.

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

Calculate -1-2i to the power of 3 and get 11+2i.

$$11+2i$$

Real Part

$$11$$

Solution Steps

Multiply both numerator and denominator of \frac{1-3i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(1-3i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$Re(\left(\frac{-2-4i}{2}\right)^{3})$$

Divide -2-4i by 2 to get -1-2i.

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

Calculate -1-2i to the power of 3 and get 11+2i.

$$Re(11+2i)$$

The real part of 11+2i is 11.

$$11$$