Solve =1+i/1-i

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

Evaluate

$$i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply complex numbers 1+i and 1+i like you multiply binomials.

$$\frac{1\times 1+i+i+i^{2}}{2}$$

By definition, i^{2} is -1.

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

Do the multiplications in 1\times 1+i+i-1.

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

Combine the real and imaginary parts in 1+i+i-1.

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

Do the additions in 1-1+\left(1+1\right)i.

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

Divide 2i by 2 to get i.

$$i$$

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.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply complex numbers 1+i and 1+i like you multiply binomials.

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

By definition, i^{2} is -1.

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

Do the multiplications in 1\times 1+i+i-1.

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

Combine the real and imaginary parts in 1+i+i-1.

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

Do the additions in 1-1+\left(1+1\right)i.

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

Divide 2i by 2 to get i.

$$Re(i)$$

The real part of i is 0.

$$0$$