Range perpendicular to Nullspace I'm stuck in this Linear Algebra problem:

Let $A\in M_n(\mathbb{C})$ with $\mathrm{rank}(A)=k$. Prove that the following are equivalent:
a) $R(A) \bot N(A)$
b) $N(A)=N(A^*)$
c) $R(A)=R(A^*)$

for a) implies b) I should prove double contention, i.e. $N(A)\subset N(A^*)$ and $N(A^*)\subset N(A)$. So for the first contention, I took $x \in R(A)$ so $\exists \ u \in \mathbb{C}^n$ such that $Au=x$, then I took $y \in N(A)$.  And by hypothesis (I mean $R(A) \bot N(A)$ ), $(Au)^*y=0$ this is the same as $u^*A^*y=0$. I had to prove that $y$ is in $N(A^*)$ i.e. $A^*y=0$ but i've got this upset $u^*$, how do I can improve that? Or from here can I conclude that $y\in N(A^*)$? Thanks in advance.

If $A^*y = u \neq 0$ then $(y,Au)=(A^*y,u)>0$, where $(.,.)$ is the inner product. However $Au \in R(A)$ and $y \in N(A)$ by assumption so it must be the case that $(y,Au)=0$, a contradiction.
