So, in essence, my professor for my complex analysis class gave us a question: essentially, we calculate...

$$\int_L |z|^2 dz$$

...along two different paths $L$ in the complex plane. These two paths:

  • The straight line from $z_0=1$ to $z_1=i$
  • The arc of the unit circle from $z_0=1$ to $z_1=i$, specifically limited to the first quadrant of the complex plane.

If we parameterize the curves, using $f(z)=|z|^2$ with a parameterization $z(t)$, then we calculate this via the integral

$$\int_L f(z(t))z'(t)dt$$

For each path $L$, I believe we get the parameterizations that follow:

  • For the first path (straight line), we use the parameterization $z(t)=(1-t)z_0+tz_1$ where $t \in [0,1]$. For our specific start/end points, then, we have the parameterization $z(t)=(1-t)(1)+t(i)=(1-t)+i(t)$
  • For the second path, we use the parameterization for the standard unit circle in the complex plane, just limited to the first quadrant. So we use $z(t)=e^{it}$ where $t \in [0,\pi/2]$ (as opposed to $[0,2\pi]$ for the whole circle).

Respectively, we have the derivatives:

  • Straight line: $z'(t)=-1+i$
  • Circle arc: $z'(t)=ie^{it}$

Also, we have the equations for $f(z(t))$ as:

  • Straight line: $f(z(t)) = | (1-t)^2 + t^2 |^2 = (1-t)^2 + t^2 = 2t^2 - 2t + 1$
  • Circle arc: $f(z(t)) = | e^{it} |^2 = 1$ (this follows from the fact that the modulus of $e^{it}$ is $1$, and in turn the square of said modulus is also $1$)

Thus, the straight line integral (the calculation here is slightly shortened but retaining the essence of what we should have):

$$\int_L f(z(t))z'(t)dt = \int_{t=0}^{t=1} (2t^2 - 2t + 1)(-1 + i)dt = (-1+i) \left( \frac{2}{3}t^3 - t^2 + t \right) |_0^1 = \frac{2}{3}(-1+i)$$

Similarly, the circle arc integral (again, slightly shortened for brevity's sake):

$$\int_L f(z(t))z'(t)dt = \int_{t=0}^{t=\pi/2} ie^{it}dt = e^{it}|_0^1 = -1+i$$

Now, to recollection, integrals are generally "path invariant" - I'm not sure of the correct terminology, but essentially all paths with the same starting and ending points should have equal integrals, provided they don't cross any discontinuities. (This hasn't been covered in our class outside of a sort of simple example - we never even provided a sort of rigorous definition for it - I'm just mostly drawing on other knowledge.)

The fact that these integrals aren't equal have suggested several possibilities to me:

  • There is some sort of discontinuity or something along a path. I don't think this is true, though.
  • Maybe the integrals aren't path independent, at least in this case? I'm not 100% on this though: the question on the homework the professor gave basically implied that the integrals should be equal. I'd sooner assume I'm wrong than him.
  • My calculations are wrong? The question then becomes where? The parameterizations I feel are right; the calculations I feel are also right, so either I'm looking over something exceptionally subtle, or, more likely, I'm just being dumb.

Any ideas or nudges in the right direction?

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    $\begingroup$ I didn't read the full question, but in general integrals are not path invariant; they are only path invariant if there is some sort of conservative quantity involved. In the complex case, this is being holomorphic; but $|z|^2$ is not holomorphic. $\endgroup$ – user296602 Sep 26 '18 at 22:12
  • $\begingroup$ Hm, I didn't know that. That's interesting; thanks. $\endgroup$ – Eevee Trainer Sep 26 '18 at 22:13

Your computions look fine. But there is a misconception there. In order that the integrals are path invariant, the function that you are integrating must have a primitive (actually, the conditions are equivalent). Your function hasn't one.

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  • $\begingroup$ Hm, that isn't something we've gone over in class. I feel like - from what I can look up - it is essentially the complex version of the antiderivative from calculus. So in essence, what you're saying is that there isn't path invariance, as a result of $|z|^2$ not having a proper antiderivative, similar to $|x|^2$ in the reals? But as a different example, $z^2$ (now sans modulus) would have invariance? $\endgroup$ – Eevee Trainer Sep 26 '18 at 22:27
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    $\begingroup$ @EeveeTrainer Yes! It would be $\frac{z^3}3$, of course. $\endgroup$ – José Carlos Santos Sep 26 '18 at 22:34

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