Solve (2i/1+i)^2

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

Evaluate

$$2i$$

Solution Steps

Multiply both numerator and denominator of \frac{2i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{2i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide 2+2i by 2 to get 1+i.

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

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

$$2i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{2i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{2i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide 2+2i by 2 to get 1+i.

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

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

$$Re(2i)$$

The real part of 2i is 0.

$$0$$