The existence of anti-derivatives

Question

The only thing I can think of is that the function is continuous hence the anti derivative exists. I was wondering if there is anything else that needs to be done/said?

Answer 1

Your idea "the function is continuous hence the anti-derivative exits" doesn't work. This is true for real functions, but it is not true anymore in the complex case.

For example, complex conjugation $z\mapsto \bar z$ is continuous, but is not the derivative of anything (on any nonempty open set).

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After a course in complex analysis, the argument that $z\mapsto e^{z^2}$ is a derivative would be as easy as

$z\mapsto e^{z^2}$ is differentiable everywhere by the chain rule. Therefore it has an antiderivative everywhere.

But if you don't yet know that a differentiable function on a simply connected domain always has an antiderivative, you may need to do more than that. What you need to do depends on what you do know so far, though.