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I have to say if this is true or not and why.

Let f a complex function, then $$Re\int_{\gamma}{f(z)dz}=\int_{\gamma}{Re(f(z))dz}$$

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closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 '18 at 4:56

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    $\begingroup$ No. ${}{}{}{}{}$ $\endgroup$ – T. Bongers Dec 7 '18 at 22:17
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    $\begingroup$ How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i \theta}$. $\endgroup$ – mlerma54 Dec 7 '18 at 22:32
  • $\begingroup$ Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $\gamma$. $\endgroup$ – zahbaz Dec 7 '18 at 23:13
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$$Re\int_{\gamma}{\frac{i}{z}\text{ } dz}=-2\pi$$

$$\int_{0}^{2\pi}-Re\frac{1}{e^{it} }dt=\int_{0}^{2\pi}-costdt=0$$

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$dz$ is complex for a general path $\gamma$. Your statement would be true if $\gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) \neq Re(a)b$. Indeed, they are equal iff $b$ is real.

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