Solve i^3/1+i+i^5

$$\frac { i ^ { 3 } } { 1 + i } + i ^ { 5 }$$

Evaluate

$$-\frac{1}{2}+\frac{1}{2}i=-0.5+0.5i$$

Solution Steps

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

$$\frac{-i}{1+i}+i^{5}$$

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

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

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

$$\frac{-1-i}{2}+i^{5}$$

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

$$-\frac{1}{2}-\frac{1}{2}i+i^{5}$$

Calculate i to the power of 5 and get i.

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

Add -\frac{1}{2}-\frac{1}{2}i and i to get -\frac{1}{2}+\frac{1}{2}i.

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

Real Part

$$-\frac{1}{2} = -0.5$$

Solution Steps

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

$$Re(\frac{-i}{1+i}+i^{5})$$

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

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

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

$$Re(\frac{-1-i}{2}+i^{5})$$

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

$$Re(-\frac{1}{2}-\frac{1}{2}i+i^{5})$$

Calculate i to the power of 5 and get i.

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

Add -\frac{1}{2}-\frac{1}{2}i and i to get -\frac{1}{2}+\frac{1}{2}i.

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

The real part of -\frac{1}{2}+\frac{1}{2}i is -\frac{1}{2}.

$$-\frac{1}{2}$$