# Is this true? $Re\int{f(z)dz}=\int{Re(f(z))dz}$ [closed]

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

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.

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

## 2 Answers

$$Re\int_{\gamma}{\frac{i}{z}\text{ } dz}=-2\pi$$

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

$$dz$$ 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.