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.

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

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

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

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

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

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

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

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

$$0$$