Show that the pumping lemmas for context-free and regular languages are equivalent for unary languages
let $ z \in L$ such that $ |z| \geq n $ (where $n$ is the lemma's constant).
By the pumping lemma for CFL, we know that we can write $z=uvwxy$ such that:


$|vwx| \leq n$
$|vx| \geq 1$
for all $i \geq 0$: $uv^iwx^iy \in L$


Because $L$ is over an unary alphabet, we can change the order of the sub-words and the word ($z$) will not change, meaning we can also write $z=wvxuy$.

So. for all $i \geq 0$, $uv^iwx^iy = wv^ix^iuy = w(vx)^iuy$.
Let's write a little different:
$u' = w, v' = vx, w' = uy$. and we have that $u'(v')^iw' \in L$.

It's easy to see that $|u'v'| \leq n$ and $|v'| \geq 1$, 
So we can conclude that for the same $n$, the conditions of pumping lemma for regular languages holds. 

It might be worth mentioning that every CFL of unary alphabet is also regular (We know that even though a language satisfies the pumping lemma for regular/CF languages, it does't mean that the language is regular/CF).

It can be shown using Parikh's theorem, and showing that for every semi-linear set $S \subseteq \mathbb{N}  $ there is a regular language $L$ such that $\Psi (L)=S$ (or $p(L)$ using wikipedia's notations)