If two sums of real power functions are equal, are the exponents equal? Suppose we have $f(z) = z^{a_1} + z^{a_2} + ... + z^{a_n} = g(z) = z^{b_1} + z^{b_2} + ... + z^{b_n}$, $z \in \Bbb C \setminus \{z \in \Bbb C \mid Rez \leq 0 \}$, $a_i, b_i \in \Bbb R^*$, $a_1 \leq a_2 \leq ... \leq a_n$, and $b_1 \leq b_2 \leq ... \leq b_n$.
Do we have $a_i = b_i \ \forall i = 1, 2, ...n$ ? 
We define the power function for real exponents with the complex exponential and logarithm : $z^{a_i} = \exp(a_i\log(z))$.
For $a_i$, $b_i \in \Bbb N$, this is certainly true, but what about the other cases ?
 A: If you have it for all $z \in \Bbb R$ it is true.  If $a_n \neq b_n$ you can take $z$ large enough that the larger one dominates the sum.  Subtract those terms off and continue, finding $a_{n-1}=b_{n-1}$, and so on.
A: You should specify a branch cut of the logarithm, I guess, otherwise you end up running into trouble. Anyway, the identity on the positive real axis suffices: if 
$$
x^{a_1}+\ldots + x^{a_n} = x^{b_1}+\ldots+ x^{b_n}$$
holds for all $x\in (0, \infty), $  then taking the limit as $x\to 0$ you see that $a_1=b_1$, therefore 
$$
x^{a_2}+\ldots + x^{a_n} = x^{b_2}+\ldots + x^{b_n}.$$
Iteratively you show that all exponent are equal. 
Note that we could have required the identity to hold only in a right neighborhood of $0$, or in a left neighborhood of $+\infty$, for this argument to work.
A: I might be wrong but let's take
Z: Real(Z)=1 , Img(Z)=0
What we get is
$1^{a1} + ... + 1^{an} = 1^{b1} + ... + 1^{bn}$.
Let's take $n=1$
Then $1^{a1} = 1^{b1}$
Which you can see clearly that $a1$ doesn't has to be equal to $b1$
example: $1^4 = 1^8$
