Consider the ring $ R = \mathbb{Z}[\frac{1 + \sqrt{-15}}{2}]$, then what are the units of this ring?
My try;
Let $\alpha \in R$. Then $\alpha = a + b(\frac{1 + \sqrt{-15}}{2})$, for some $a,b \in \mathbb{Z}$.
Now, $\alpha$ is a unity iff $N(\alpha) = 1$ iff $(a + b(\frac{1 + \sqrt{-15}}{2}))(a - b(\frac{1 + \sqrt{-15}}{2})) = 1$ iff $a^2 - b^2 (\frac{1 + \sqrt{-15}}{2})^2 = 1$.
Now, from here, can I simply state that $b = 0$ since we need to get rid of the complex part of the LHS? Then deduce that $\alpha = 1$ or $\alpha = -1$?
In general $\alpha$ is a unit in $R$ if and only if $N(\alpha)$ is a unit in $\Bbb{Z}$. You haven't considered the case $N(\alpha)=-1$.
Second, the norm of $\alpha=a+b\left(\tfrac{1 + \sqrt{-15}}{2}\right)$ is not equal to $$N(\alpha)=\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a - b\left(\frac{1 + \sqrt{-15}}{2}\right)\right).$$ Instead, it is given by taking the product over the conjugates of $a+b\left(\tfrac{1 + \sqrt{-15}}{2}\right)$, which gives $$N(\alpha)=\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a+ b\left(\frac{1 - \sqrt{-15}}{2}\right)\right),\tag{1}$$ because $\frac{1-\sqrt{-15}}{2}$ is the other root of the minimal polynomial of $\frac{1+\sqrt{-15}}{2}$.
Proceeding from $(1)$ you want to find all integral solutions to $$\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a+ b\left(\frac{1 - \sqrt{-15}}{2}\right)\right)=\pm1.$$ Expanding the left hand side of the equation yields an equation over the integers, and its solutions correspond to the units.