Morera's Theorem Proof 
Hello, I am having trouble with Morera's Theorem. How does the integral being equal to 0 matter? I can't see why this condition is necessary for this theorem to hold true. Also, if someone could provide an example of how to show an arbitrary closed curve is equal to 0 without the use of Cauchy's theorem would be nice. 
 A: The fact that the integral along any closed curve is 0 allows us to unambiguously define the integral $∫_a^z f(z) \ dz$, which is independent of the path from $a$ to $z$ used. (note your notes says 'over any curve')
To see this, say we have 2 paths $p_1, p_2$ both starting at $a$ and ending at $z$. Then the path $p$ defined to be $p_1$ followed by $p_2$ in reverse is a closed curve, so
$$ ∫_p f(z) dz = 0 $$ by assumption. But this is precisely saying that
$$∫_{p_1} f(z) dz - ∫_{p_2} f(z) dz  = 0$$
Alternatively, a non-analytic function will not satisfy this criteria, so it is a necessary condition. For instance take $f(z) = \bar{z}$, then
$∫_{-1}^1 f(z) dz = 0$ via the path in $\mathbb{R}$, but if we instead use the path defined by $\gamma(t) := e^{π i t}, t∈ [1,2]$ then we get
$$ ∫_\gamma f(z) dz = ∫_1^2 e^{-π i t} \ π i e^{π i t}d t  = π i$$
A: What book is this, I wonder? I would say this: If $A$ is any open subset of $\mathbb {C},$    $\,f: A \to \mathbb {C}$ is continuous, and $\int_\gamma f(z)\,dz = 0$ for every closed contour $\gamma $ in $A,$ then $f$ is holomorhic in $A.$ "Simply connected" has nothing to do with Morera.
