If A is NL-complete then complement of A is also NL-complete?
Start with the definition of $coNL$.
$$coNL = \{A\mid \overline{A} \in NL\}$$
Since $NL=coNL$, every $coNL$-complete language is $NL$-complete as well.
Let $A$ be a $NL$-complete problem. Then 
$$\forall B \in NL,B \leq_L A$$
or equivalently 
$$x \in B \Longleftrightarrow f(x) \in A, \text{  where } f \text{is logspace computable}$$
$$x \notin B \Longleftrightarrow f(x) \notin A, \text{  where } f \text{is logspace computable}$$ 
$$x \in \overline{B} \Longleftrightarrow f(x) \in \overline{A}, \text{  where } f \text{is logspace computable}$$ 

and so $$\forall \overline{B} \in coNL,\overline{B} \leq_L \overline{A}$$
This shows that $\overline{A} $ is $coNL$-complete and hence $NL$-complete (since $coNL=NL)$.