Multiplicative inverse of a complex number.

Let $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ be the inverse of $1+3\sqrt{2}$, i.e. $(a+b\sqrt{2})(1+3\sqrt{2})=1$. Then $$ 1=a+6b+(3a+b)\sqrt{2}, $$ i.e. $$ 3a+b=0,\ a+6b=1. $$ It follows that $$ a=-\frac{1}{17},\ b=\frac{3}{17}. $$ Now $$ (1+3\sqrt{2})x=1-5\sqrt{2} \iff x=(1+3\sqrt{2})^{-1}(1-5\sqrt{2}), $$ i.e. $$ x=\frac{1}{17}(-1+3\sqrt{2})(1-5\sqrt{2})=-\frac{31}{17}+\frac{8}{17}\sqrt{2}. $$

The inverse is almost never the conjugate. However, it does end up being related to the conjugate. (Why and how?) We can also use the conjugate instead, and avoid having to determine the inverse explicitly. Multiplying both sides of $$(1+3\sqrt{2})x=1-5\sqrt{2}$$ by $1-3\sqrt{2}$ gives us $$-17x=31-8\sqrt{2},$$ from which we see that $$x=-\frac{31}{17}+\frac8{17}\sqrt{2}.$$

Hint $\ $ If $\rm\: 0\ne\alpha\bar\alpha = n\in \Bbb Z\:$ then $\rm\: \alpha\, x = \beta\:$ times $\,\bar \alpha\,$ yields $\rm\: n\, x = \bar\alpha\beta, \:$ i.e. in fraction language $$\rm\: x = \dfrac{\beta}\alpha = \dfrac{\bar\alpha\beta}{\bar\alpha\alpha} = \dfrac{\bar\alpha\beta}n$$

This is known as rationalizing the denominator. It exploits the fact that every irrational algebraic $\alpha$ has a "simpler" (i.e. rational) multiple (its norm), to reduce division by an irrational to division by a rational. This is a special case of the general method of simpler multiples.