Solve (1+i)^4/1-i+2

$$\frac { ( 1 + i ) ^ { 4 } } { 1 - i } + 2$$

Evaluate

$$-2i$$

Solution Steps

Calculate 1+i to the power of 4 and get -4.

$$\frac{-4}{1-i}+2$$

Multiply both numerator and denominator of \frac{-4}{1-i} by the complex conjugate of the denominator, 1+i.

$$\frac{-4\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+2$$

Do the multiplications in \frac{-4\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$\frac{-4-4i}{2}+2$$

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

$$-2-2i+2$$

Add -2-2i and 2 to get -2i.

$$-2i$$

Real Part

$$0$$

Solution Steps

Calculate 1+i to the power of 4 and get -4.

$$Re(\frac{-4}{1-i}+2)$$

Multiply both numerator and denominator of \frac{-4}{1-i} by the complex conjugate of the denominator, 1+i.

$$Re(\frac{-4\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+2)$$

Do the multiplications in \frac{-4\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$Re(\frac{-4-4i}{2}+2)$$

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

$$Re(-2-2i+2)$$

Add -2-2i and 2 to get -2i.

$$Re(-2i)$$

The real part of -2i is 0.

$$0$$