Solve (1/1+i)^2

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

Evaluate

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

Solution Steps

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

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

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

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

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

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

Calculate \frac{1}{2}-\frac{1}{2}i to the power of 2 and get -\frac{1}{2}i.

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

Real Part

$$0$$

Solution Steps

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

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

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

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

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

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

Calculate \frac{1}{2}-\frac{1}{2}i to the power of 2 and get -\frac{1}{2}i.

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

The real part of -\frac{1}{2}i is 0.

$$0$$