We have integral: $$\int_0^1\exp\left(n\left(\frac{itz}{\sqrt{a(1-a)n}}+a\ln(z)+(1-a)\ln(1-z)\right)\right)dz=\int_0^1\exp(nf(z))dz,$$ where $0<a<1$.
We want to approximate this integral when $n\rightarrow\infty$ by method of steepest descent. Why deforming $[0,1]$ to the contour through saddle point has the following expansion: $$z=a+\frac{it\sqrt{a(1-a)}}{\sqrt{n}}+O\left(\frac{1}{n}\right)?$$ I found saddle point $z_0=a$ from $\text{Re}'(f(z))=0$. I don't know what to do next? I can’t understand how they got it? I hope for your help! Thank you!
my advice would be to try rewriting your integral as $$ I = \int_0^1 z^{an}(1-z)^{n(1-a)} \exp\left({\sqrt{n}xiz}\right) dz, $$ where $x(t,a):=\frac{t}{\sqrt{a(1-a)}}.$
Then there are no stationary points, but you have a simple linear phase, so steepest descent can be applied easily by deforming onto the vertical contours Re(z)=0 and Re(z)=1, like so $$ I = i\int_0^\infty (ip/x)^{an}(1-p/x)^{n(1-a)} \exp\left(-np\right) dp -ie^{\sqrt{n}xi}\int_0^\infty (1+ip/x)^{an}(p/x)^{n(1-a)} \exp\left(-np\right) dp $$ where the first and second integral correspond to the integral up Re(z)=0 and down Re(z)=1 respectively.
At this point, you could binomially expand the (1-p/x)^{n(1-a)} and (1+ip/x)^{an} terms, aiming for an asymptotic expansion of the integrand which holds for $p\to0$. Once you've done this, apply Watson's Lemma to both integrals.
I hope this helps :-)