finite dimensional algebras without bound quiver representation

(1) For the $\mathbb{R}$-algebra $\mathbb{C}$, ($\mathbb{R}$ real number field, $\mathbb{C}$ complex number field) there is no quiver $Q$ such that $\mathbb{C}\cong \mathbb RQ/\mathcal{I}$ with $\mathcal{I}$ an admissible ideal of $\mathbb{R}Q$, why?

(2) Let $A$ be the $\mathbb{R}$-algebra$\left[ \begin{array}{cc} \mathbb{C} & \mathbb{C} \\ 0 & \mathbb{R} \\ \end{array} \right]$. Then $A$ is a basic $\mathbb{R}$-algebra, but there is no quiver $Q$ such that $A\cong \mathbb{R}Q/\mathcal{I}$ with $\mathcal{I}$ an admissible ideal of $\mathbb{R}Q$. why?

-

I can answer question (1) - $\mathbb{C}$ has exactly one (non-zero) idempotent, and zero radical, so the quiver $Q$ in question must have exactly one vertex, and no arrows. Then there are no non-zero admissible ideals, but $\mathbb{R}Q\cong\mathbb{R}$.
I haven't worked it all the way through, but I imagine a similar approach works for the second question - find the idempotents and the radical to write down what $Q$ has to be, and then try to understand why no possible ideals give you the right quotient.