Solve 1+1/i/1-frac{1{i}}

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

Evaluate

$$-i$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

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

Divide i by -1 to get -i.

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

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

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

Divide i by -1 to get -i.

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

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

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

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

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

Divide -2i by 2 to get -i.

$$-i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

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

Divide i by -1 to get -i.

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

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

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

Divide i by -1 to get -i.

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

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

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

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

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

Divide -2i by 2 to get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$