Proposition (holomorphic function has Taylor expansion convergent on its domain):
Let be open, and let be holomorphic. Let . Then there exists coefficients of a Taylor series such that whenever is a ball of radius that is contained within , then the Taylor series
converges absolutely on and equals there.
Proof: By Cauchy's formula, we have
Adding a zero, we may rewrite this as
Computing further, and using the convergence of the geometric series for arguments of modulus ,
here, interchanging differentiation and integration is justified by the absolute convergence of the former expression.
Theorem (identiy theorem):
Let be open, and let be two holomorphic functions defined on . Suppose that the point is an accumulation point of the set . Then for all in the connected component of of .
Proof: We develop both
Theorem (Riemann's theorem on removable singularities):
Let be a function which is holomorphic on for a certain , and furthermore bounded in a ball (or rather disk) about , say , where is just the radius of that small ball. Then we will find a value such that if we continue to at the point , the result will be holomorphic.
Proof: We define the new function
and claim that it's holomorphic on . It will be complex differentiable in "by the product rule" (which is supposed here to include the statement that the product of complex differentiable functions is complex differentiable), and in we have
since is bounded.
Now we develop into a Taylor series at . But the first two coefficients of it will be zero, whence will be divisible by to obtain a power series, and this power series will define a holomorphic function about . But by the definition of , this series coincides with everywhere except , whence if is continued at by the constant term of the series, the result will be holomorphic and thus the desired continuation.