Inverse of $2 − 3i \text{ in } \Bbb Q(i).$

Question

Find the inverse of the element in the given field. The field is a finite extension F(α). Express your answer in the form $a_0 +a_1α+ ···+a_{n−1}α^{n−1}$, where $a_i$ ∈ F and [F(α):F]=n.

$$2 − 3i \text{ in } \Bbb Q(i).$$

I'm a bit confused on how to start this problem. I understand the terminology and notation but I don't know how to find the inverse of $2-3i$ in $\Bbb Q(i)$. Any help would be great, thank you in advance!

Answer 1

Since $i$ has degree $2$ over $\mathbb{Q}$, you know that $$ (2-3i)^{-1}=a+bi $$ for some $a,b\in\mathbb{Q}$.

Can you expand $(a+bi)(2-3i)=1$ and finish?

Answer 2

Completing the hints given to you in the comments to your question:

$$\frac{1}{2-3i}=\frac{2+3i}{(2-3i)(2+3i)}=\frac{2+3i}{13}=\frac{2}{13}+\frac{3}{13}i.$$

In general, the same technique will show you:

$$(a\pm bi)^{-1}=\frac{a}{a^2+b^2} \mp \frac{b}{a^2+b^2}i.$$