Proof of the Inverse Function Theorem in Complex Analysis
Theorem 1
A function that is analytic at and satisfies exists in the region , where exists.
Explanation
Consider the condition given by .
When thought of as a real function, it implies the function is either increasing or decreasing, which is a condition for the existence of an inverse function. Geometrically, this refers to a Smooth function, indicating there are no abrupt changes in direction or bends. What must be noted in the inverse function theorem is that even if such conditions are met, it doesn’t mean the inverse function itself exists universally but rather within a local limit.
Proof
It suffices to show that equation has a unique solution for .
If we set , then and . This means that is a simple zero of , and there exists that satisfies in .
For circle , define that satisfies . Then, the following holds in :
Rouché’s Theorem: If and are analytic on the simple closed path and satisfy on , then and have the same number of zeros inside .
By Rouché’s Theorem, and have the same number of zeros inside .
However, as seen before, since has only a single simple zero , the zero satisfying is also unique inside . Therefore, it can be concluded that equation has a unique solution inside .
If we now let , it means that has a unique solution inside at .
See Also
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Osborne (1999). Complex variables and their applications: p193. ↩︎