Solve 1+i/1-i-1-i/1+i

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

Evaluate

$$2i$$

Solution Steps

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

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

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

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

Divide 2i by 2 to get i.

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

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

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

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

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

Divide -2i by 2 to get -i.

$$i+i$$

Add i and i to get 2i.

$$2i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

Divide 2i by 2 to get i.

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

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

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

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

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

Divide -2i by 2 to get -i.

$$Re(i+i)$$

Add i and i to get 2i.

$$Re(2i)$$

The real part of 2i is 0.

$$0$$