Now that addition and multiplication of complex numbers are defined, one can go on to show that the complex numbers form a field. Many of the properties are easy to show. Here, we show what the additive and multiplicative inverses of complex numbers are. They will then give us subtraction and division.

In what follows, \(z = a+bi\) and \(w = c+di\) are complex numbers with \(a,b,c,d\in \mathbb{R}\).

The additive inverse of \(z\), denoted by \(-z\), is \((-a)+(-b)i\), or more simply, \(-a-bi\). For example, \( -(1+2i) = -1-2i\). Clearly, \(z + (-z) = 0\) since \( (a +bi) + (-a-bi) = (a-a) + (b-b)i = 0 + 0i = 0\).

Subtraction is then defined as follows: \[z-w = z + (-w).\] For example, \( (2+i) - (1+i) = (2+i) + ((-1) + (-1)i) = 1\) since the additive inverse of \(1+i\) is \((-1) + (-1)i\). Once one has understood the definition of subtraction, one could write in a less pedantic manner: \[ (2+i) - (1+i) = 2+i - 1 - i = 1.\]Suppose that \(z \neq 0\). Is there a multiplicative inverse of \(z\)? If so, what is it?

Note that \(z \neq 0\) implies that not both \(a\) and \(b\) are \(0\). We want to see if we can find real numbers \(c\) and \(d\) so that \(z w = 1\), or equivalently, \[(ac-bd) + (ad+bc)i = 1.\]

As the right-hand side is a real number, we need the imaginary part on the left-hand side to be zero: \[ad+bc = 0.\] That leaves \[ac-bd = 1.\] Note that \(c = \frac{a}{a^2+b^2} \text{ and } d = -\frac{b}{a^2+b^2}\) satisfy the two equations above. (These were obtained by solving for \(c\) and \(d\) in terms of \(a\) and \(b\). The details are quite messy and are omitted here.)

Indeed, \[ad + bc = a\cdot\left(-\frac{b}{a^2+b^2}\right) + b\cdot \frac{a}{a^2+b^2} = -\frac{ab}{a^2+b^2} +\frac{ab}{a^2+b^2} = 0\] and \[ac - bd = a\cdot\frac{a}{a^2+b^2} + -b\cdot\left(- \frac{a}{a^2+b^2}\right) = \frac{a^2}{a^2+b^2} + \frac{b^2}{a^2+b^2} = \frac{a^2+b^2}{a^2+b^2} = 1.\]

Since not both \(a\) and \(b\) are \(0\), \(a^2+b^2 \neq 0\) and so \(c\) and \(d\) are well-defined real numbers. Hence, the multiplicative inverse of \(z\) is \[z^{-1} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2} i.\]

Note that \(i (-i) = - i^2 = - (-1) = 1\). Hence, \(i^{-1} = -i\).

What is the multiplicative inverse of \(1-2i\)? Using the formula above with \(a = 1\) and \(b = -2\), we get \[(1-2i)^{-1} = \frac{1}{1^2+(-2)^2} - \frac{-2}{1^2+(-2)^2} i = \frac{1}{5} + \frac{2}{5}i.\] One can check indeed that \((1-2i)\left(\frac{1}{5} + \frac{2}{5}i\right)= 1\).

Dividing \(z\) by \(w\), denoted by \(\displaystyle\frac{z}{w}\) or \(z/w\), is given by \(z\) multiplied by the multiplicative inverse of \(w\). In other words, \[\frac{z}{w} = z w^{-1}.\]

What is \(\displaystyle\frac{i}{1+i}\)?

Let's first find the multiplicative inverse of \(1+i\). Using the formula for multiplicative inverse, we get \((1+i)^{-1} = \frac{1}{2}- \frac{1}{2}i\). Hence, \[\begin{eqnarray} \frac{i}{1+i} & = & i (1+i)^{-1} \\ &= & i \left( \frac{1}{2} - \frac{1}{2}i\right) \\ &= & \frac{1}{2}i - \frac{1}{2} i^2 \\ &= & \frac{1}{2}i + \frac{1}{2} \\ &= & \frac{1}{2} + \frac{1}{2}i \end{eqnarray}\]

Let \(z = 1-2i\) and \(w = -2+i\). Give the answer to each of the following. Express your answer in the form \(a+bi\) with \(a,b\in \mathbb{R}\).

\(z - w\)

\(\displaystyle\frac{z}{w}\)

\(\displaystyle\frac{w}{z}\)