i was ask to prove that
$$\int_{-1}^{1} Re(z) dz$$ is not well defined.
My attemp:
I want to prove that there is two countor $C_1$ and $C_2$ that connect -1 and 1 such that $$\int_{C_1} Re(z) dz \not =\int_{C_1} Re(z) dz$$
Let $C_1$: $[-1,1] \to \mathbb{C}$ defined as follow $t \mapsto t$
Adn $C_2$: $[0,\pi] \to \mathbb{C}$ defined as follow $t \mapsto e^{it}$
So $$\int_{C_1} Re(z) dz= \int_{-1}^{1} t dt=0$$
and $$\int_{C_2} Re(z) dz= \int_{0}^{\pi} e^{t}\cdot ie^{it}= i \int_{0}^{\pi}e^{t(1+i)}= \dfrac{e^o}{1+i}-\dfrac{e^{\pi(1+i}}{1+i}\not =0$$
Is my answer ok?
The reason for which the integral is that the real part of $z$ is nowhere analytical.
However, I am afraid that you have reached a correct conclusion for a wrong reason. $$ \int_{C_2} \operatorname{Re} {(z)} \,\mathrm{d}z = \int_{0}^{2\pi/2} \operatorname{Re} {(\mathrm{e}^{\mathrm{i}t})}\, \mathrm{i} \mathrm{e}^{\mathrm{i}t} \,\mathrm{d}t = \int_{0}^{2\pi/2} (-\sin t \cos t + \mathrm{i} \cos^2 t) \,\mathrm{d}t = \frac{2\pi \mathrm{i}}{4}. $$