It is common that De Moivre's formula works when n belongs to rational number. $$(\cos a + i \sin a)^n= \cos na + i \sin na $$
Can this formula works when n belongs to complex number? I tried to prove them by connection between rational number and complex number but I failed.
It does, but there will be other values as well. In fact, unless $n$ is an integer, there are always multiple values (for example, if $n=1/2$, there are two values which are opposites). Note that your statement for the case of rational $n$ is not quite correct.
The general rule is $z^w = e^{w\log z}$ which is typically multiple-valued because $\log z$ is multiple-valued. In your case, regard $z = \cos a + i\sin a = e^{ia}$, so that the values of $\log z$ are $(a + 2k\pi) i$ (for integral $k$). Then the rule would say that the values you want are $$e^{n\log z} = e^{n(a + 2k\pi )i} = \cos (na + 2kn\pi ) + i\sin (na + 2kn\pi )$$
If $n$ is an integer, these are the same for all $k$ and are $\cos na + i\sin na$.
If $n$ is rational but not an integer, there are a finite number of different values.
If $n$ is irrational or not real, there are a countably infinite number of different values.
The case where $n$ is not real can be broken down a little more in terms of real and imaginary parts using the facts that $$\cos(x+iy)=\cos x \cosh y - i \sin x \sinh y$$ and $$\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$$
Addendum: The easiest form to use is when $z$ is written as $re^{it}$ and the exponent $w$ is written as $a+bi$ (with the usual assumption that $r>0$ and all of $r, t, a, b$ are real). Then you have (with the convention that "$\ln$" is the ordinary real-valued logarithm of a positive real number)
$$\left(re^{it}\right)^{a+bi} = e^{(a+bi)\log(re^{it})}$$ $$=e^{(a+bi)(\ln r + (t+2k\pi)i)}$$ $$=e^{a\ln r -(t+2k\pi)b + i(b\ln r+(t+2k\pi)a)}$$ $$=e^{a\ln r -(t+2k\pi)b}\cdot e^{i(b\ln r+(t+2k\pi)a)}$$ $$=r^ae^{(t+2k\pi)b}\cdot e^{i(\ln r^b + (t+2k\pi)a)}$$ which, if desired, can be written as $$r^ae^{(t+2k\pi)b}\cos(\ln r^b + (t+2k\pi)a) + ir^ae^{(t+2k\pi)b}\sin(\ln r^b + (t+2k\pi)a)$$
The "base" value is when $k=0$: $$r^ae^{tb}\cdot e^{i(\ln r^b + ta)} = r^ae^{tb}\cos(\ln r^b + ta) + ir^ae^{tb}\sin(\ln r^b + ta)$$
Finally, note that when $b=0$, the exponent is real and we have $$r^a\cdot e^{(t+2k\pi)ai} = r^a\cos(t+2k\pi)a + ir^a\sin(t+2k\pi)a$$ which yields your familiar DeMoivre form when $k=0$: $$r^a\cdot e^{tai} = r^a\cos ta + ir^a\sin ta$$