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.

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

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

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

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

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

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.

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

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

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

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

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

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.

$$0$$