# The complex conjugate squared has no antiderivative

I am aware of the theorem that a function in the complex plane $f(z)$ has no antiderivative if and only if the contour integral of $f(z)$ over every closed path is non-zero. Because it is also a theorem that each closed path is contractable to a circle, therefore this is the same as integrating it over ever possible circle.

However, the integral is zero (so it's supposed to have an anti-derivative) whenever the circle is centered at the origin. But I know that this function never has an antiderivative. So what went awry?

• Your question is not very clear. Please state clearly in the body of your question and not just the title what you are asking. I think you may be asking why Morera's theorem does not imply that $z \mapsto (z^*)^2$ is holomorphic. May 29, 2018 at 21:11

Because you can't deduce frm the fact that one path $\gamma$ is contractible to another path $\gamma^\star$ that the integral of a functions along $\gamma$ is equal to the integral of that function along $\gamma^\star$. That is true for holomorphic functions, but that's not the case here.