$\oint_C\text{Re}(z)dz$ is independent of the path $C$ between $z_0 = 0$ and $z_1 = 1+i$.

Question

My way to prove that$\oint_C\text{Re}(z)dz$ is independent of the path $C$ between $z_0 = 0$ and $z_1 = 1+i$. This statement is false. I tried to take $z(t)=t+it$ and then I found this; $\oint_C〖(t+it)dt〗=(1/2)(t^2/2+it^2/2)$ But I couldn't keep move. How can I prove this statement is false? What should I do to keep moving? If I take another $z_1(t)=t^2+it$ (and that is the different path) can I prove this statement is false like that?

Answer 1

To prove the statement false you only need to find two different contours that given different values for the integral.

Consider $ \gamma_1 $ as the path from $ 0 $ to $ 1 + i $ that travels along the real axis to $ 1 $ and then parallel to the imaginary axis to $ 1 + i $. Along the first leg, let $ z=x $ so that $ \Re z $ is simply $ x $ and along the second leg let $z = 1 +iy $ so that $ \Re z = 1 $

$$ \int_{\gamma_1} z dz = \int_0^1 x \cdot \frac{d}{dx} (x +i0) dx + \int_0^1 1\cdot \frac{d}{dy} ( 1 + iy) dy = \frac{1}{2} + i. $$

Then repeat with contour $ \gamma_2 $, first along the imaginary axis, then the real axis. The calulcation is very similar, and along the first leg set $ z = iy $ so that $\Re z = 0$, along the second set $z = x + i$ so that $\Re z = x $, $$ \int_{\gamma_2} z dx = \int_0^1 0 \cdot \frac{d}{dy}(0 + iy) dy + \int_0^1 x \cdot \frac{d}{dx}(x+i) dx = \frac{1}{2} $$

Because the two answers are different, the statement is false.