Solve (left(1+2iright)i^5+{left(1-i/1+iright)}^3)

$$( \left( 1+2i \right) { i }^{ 5 } + { \left( \frac{ 1-i }{ 1+i } \right) }^{ 3 } )$$

Evaluate

$$-2+2i$$

Solution Steps

Calculate i to the power of 5 and get i.

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

Multiply 1+2i and i to get -2+i.

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

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

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

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

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

Divide -2i by 2 to get -i.

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

Calculate -i to the power of 3 and get i.

$$-2+i+i$$

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

$$-2+2i$$

Real Part

$$-2$$

Solution Steps

Calculate i to the power of 5 and get i.

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

Multiply 1+2i and i to get -2+i.

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

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

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

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

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

Divide -2i by 2 to get -i.

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

Calculate -i to the power of 3 and get i.

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

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

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

The real part of -2+2i is -2.

$$-2$$