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

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

Evaluate

$$2i$$

Solution Steps

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

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

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

Subtract 1 from 1 to get 0.

$$2i+\left(\frac{0}{1+i}\right)^{2}$$

Zero divided by any non-zero number gives zero.

$$2i+0^{2}$$

Calculate 0 to the power of 2 and get 0.

$$2i+0$$

Anything plus zero gives itself.

$$2i$$

Real Part

$$0$$

Solution Steps

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

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

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

Subtract 1 from 1 to get 0.

$$Re(2i+\left(\frac{0}{1+i}\right)^{2})$$

Zero divided by any non-zero number gives zero.

$$Re(2i+0^{2})$$

Calculate 0 to the power of 2 and get 0.

$$Re(2i+0)$$

Anything plus zero gives itself.

$$Re(2i)$$

The real part of 2i is 0.

$$0$$