If $\int\limits_\gamma f(z)dz\in\Bbb R$ does it imply $f(z)\in \Bbb R~\forall z\in\gamma$?
This is from a proof where $\gamma$ is a circle and we have $1={1\over 2\pi}\int\limits_\gamma f(z)dz={1\over 2\pi}\int\limits_\gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.
Clearily $0=\int\limits_0^{2\pi}ie^{it}dt$ and $ie^{it}\notin \Bbb R$...
On
A you talking about a curve Integral?
If yes then f is probably a One-Form: Integration over a curve $\gamma : [a,b] \to \mathbb{R}^n$ is defined as follows:
$$ \int_\gamma f(z) dz=\int_a^b f(\gamma(z))\gamma'(z) dz $$
(I know this should be a comment rather than an answer but i don't have sufficient points jet)
$$\frac1{2\pi} \int_\gamma f(z) \ \mathrm dz = 1$$
Comparing real parts: $$\Re \left( \frac1{2\pi} \int_\gamma f(z) \ \mathrm dz \right) = 1$$
Taking real parts commutes with everything: $$1 = \Re \left( \frac1{2\pi} \int_\gamma f(z) \ \mathrm dz \right) = \frac1{2\pi} \Re \left( \int_\gamma f(z) \ \mathrm dz \right) = \frac1{2\pi} \int_\gamma \Re \left( f(z) \right) \ \mathrm dz$$
Notice, though, that $\displaystyle \int_\gamma \Re \left( f(z) \right) \ \mathrm dz = \int_\gamma f(z) \ \mathrm dz$ does not imply $\Re \left( f(z) \right) = f(z)$.