Solve (1+texttt{i})(1+1/texttt{i})

$$(1+ \texttt{i} )(1+ \frac{ 1 }{ \texttt{i} } )$$

Evaluate

$$2$$

Solution Steps

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

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

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

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

Divide i by -1 to get -i.

$$\left(1+i\right)\left(1-i\right)$$

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

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

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

$$1\times 1+1\left(-i\right)+i-\left(-1\right)$$

Do the multiplications.

$$1-i+i+1$$

Combine the real and imaginary parts.

$$1+1+\left(-1+1\right)i$$

Do the additions.

$$2$$

Real Part

$$2$$

Solution Steps

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

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

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

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

Divide i by -1 to get -i.

$$Re(\left(1+i\right)\left(1-i\right))$$

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

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

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

$$Re(1\times 1+1\left(-i\right)+i-\left(-1\right))$$

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

$$Re(1-i+i+1)$$

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

$$Re(1+1+\left(-1+1\right)i)$$

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

$$Re(2)$$

The real part of 2 is 2.

$$2$$