In the ring $\mathbb{Z}[i]\space$, which of the following elements belongs to the principal ideal $(3+i)$?
(a) $1+8i$
(b) $1+5i$
(c) $1+6i$
(d) $1+7i$
In the ring $\mathbb{Z}[i]\space$, which of the following elements belongs to the principal ideal $(3+i)$?
(a) $1+8i$
(b) $1+5i$
(c) $1+6i$
(d) $1+7i$
$$\begin{align}3\!+\!i\mid 1\!+\!ni\iff\ &\dfrac{1\!+\!ni}{\color{#c00}{3\!+\!i}} = \overbrace{\dfrac{(1\!+\!ni)(3\!-\!i)}{\ \ (3\!+\!i)(3\!-\!i)}}^{\text {rationalize denom}} =\dfrac{n\!+\!3+(3n\!-\!1)i}{\color{#0a0}{10}}\in \Bbb Z[i]\\[.2em] \iff\ & 10\mid n\!+\!3,\, 3n\!-\!1\\[.2em] \iff\ &n\equiv -3\equiv 7\!\!\!\pmod{\!10}\end{align}\qquad$$
Remark $ $ Rationalizing the denominator is an instance of the method of simpler multiples i.e. it simplifies division by an $\rm\color{#c00}{algebraic}$ integer $\,w\in\Bbb Z[i]\,$ to division by its norm multiple $\,w\bar w,\,$ which, being a $\rm\color{#0a0}{rational}$ integer $\in\Bbb Z,\,$ is simpler to divide by.
You are looking for complex numbers $a+bi$ with $a,b\in\Bbb Z$ such that there exists another one $c+di$ so $a+bi=(c+di)(3+i)=(3c-d)+(3d+c)i$. Let's see if this is possible for your first case:
a) $1+8i=(3c-d)+(3d+c)i\Rightarrow\begin{cases} 1=3c-d\\8=3d+c\end{cases}\Rightarrow\begin{cases}d=3c-1\\3(3c-1)+c=8\end{cases}\Rightarrow\begin{cases}d=3c-1\\10c=11\end{cases}\Rightarrow\begin{cases}d=3c-1\\c=\frac{11}{10}\notin\Bbb Z\end{cases}$
and at this point we don't care what's $d$ because we needed both $c$ and $d$ to be in $\Bbb Z$. Therefore $1+8i$ isn't in the principal ideal $(3+i)$.
Can you do the rest?