Showing a complex integral has zero real part

Question

I need to show that $\int_{\varphi}^{\space} \bar{z} \space dz $ has zero real part, for all smooth closed paths $\varphi$.

I've tested this with the example $\int_{C(0;1)}^{\space} \bar{z} \space dz $ (the disc centered around 0 with radius 1) and I get $2\pi i$. I can also prove that the real part is zero for all $\varphi$ being a disc.

A hand wavey proof might be to say that all closed paths may be represented as the union of discs... maybe?

I think a better approach might be to write $\varphi = u + iv$ and express $Re(\int_{\varphi}^{\space} \bar{z} \space dz)$ as a real integral, as someone suggested to me but I'm not 100% sure how exactly to do this.

Any help would be hugely appreciated.

Thanks.

Answer 1

$$Re(\int_{\varphi}^{\space} \bar{z} \space dz)=Re(\int_{\varphi} \overline{\varphi_1 (t) +i\varphi_2 (t)} (\varphi_1' (t) +i\varphi_2' (t)) dt =\int_{\varphi}( \varphi_1 (t)\varphi_1' (t) +\varphi_2 (t) \varphi_2' (t)) dt =\frac{\varphi_1^2 (t) +\varphi_2^2 (t) }{2}|_{t_0}^{t_0} =0$$