$$\mathbb Z[i]/(1+2i)\simeq\mathbb Z_5$$  Help me with this.
The reasoning my book uses is as follows:   Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.   I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$  is $10$. If someone could please explain this I would greatly appreciate it.   Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.
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