Click here to purchase the entire book in PDF format. Multiplicative Inverse Similarly to additive inverses, every number has a matching number, which, when the two are multiplied, equals 1. The only exception to this is the number 0, so, if $x$ does not equal 0, then the multiplicative inverse of $x$ is $\frac{1}{x}$ because $x * \frac{1}{x} = * \frac{x}{x} = 1$. Some books write this as $x * x^{-1} = 1$ because $x^{-1} = \frac{1}{x}$. In the case of complex math, things are unfortunately a little different because 1 divided by a complex number is... well... complex. if (a + jb) is not equal to 0 then: $$\frac{1}{a + jb} = \frac{a - jb}{a^{2} + b^{2}}$$ (2.42) We won't worry too much about how that's calculated, but we can establish that it's true by doing the following: $$(a + jb) * \frac{a - jb}{a^{2} + b^{2}} = \frac{(a + jb)(a - jb)}{a^{2} + b^{2}}$$ (2.43) $$= \frac{a^{2} + b^{2}}{a^{2} + b^{2}}$$ (2.44) $$= 1$$ (2.45) There's one interesting thing that results from this rule. What if we looked for the multiplicative inverse of $j$? In other words, what is $\frac{1}{j}$? Well, let's use the rule above and plug in $(0 + 1j)$. $$\frac{1}{a + jb} = \frac{a - jb}{a^{2} + b^{2}}$$ (2.46) $$\frac{1}{0 + j1} = \frac{0 - j1}{0^{2} + 1^{2}}$$ (2.47) $$\frac{1}{j} = \frac{-1j}{1}$$ (2.48) $$\frac{1}{j} = -j$$ (2.49) Weird, but true.