Integrate f(z) = Re(z) along contour of the square

[Shackleford]

Homework Statement

Integrate f(z) = Re(z) = \frac{z+\bar{z}}{2} along the contour of the square {z = x + iy | |x| ≤ 1, |y| ≤ 1} with counterclockwise rotation.

Homework Equations

C_j: φ_j(t) = z₀ + (z₁ - z₀)t

C₁: 1 -i + 2it
C₂: 1 +i - 2t
C₃: -1 +i - 2it
C₄: -1 -i + 2t

The Attempt at a Solution

If I'm not mistaken, because of linearity, I can simply integrate the real parts of parametrizations.

 

[Ray Vickson]

Homework Statement

Integrate f(z) = Re(z) = \frac{z+z^bar}{2} along the contour of the square {z = x + iy | |x| ≤ 1, |y| ≤ 1} with counterclockwise rotation.

Homework Equations

C_j: φ_j(t) = z₀ + (z₁ - z₀)t

C₁: 1 -i + 2it
C₂: 1 +i - 2t
C₃: -1 +i - 2it
C₄: -1 -i + 2t

The Attempt at a Solution

If I'm not mistaken, because of linearity, I can simply integrate the real parts of parametrizations.

Try it and see what you get.

For TeX/LaTex commands: the command to get $\bar{z}$ is "\ b a r {z}" (no spaces). Also, the command "a^ bc " (no spaces) produces $a^bc$. If you want more than one symbol in a superscript or a subscript, use { }, like this: 'a ^ {b c}" no spaces; that gives $a^{bc}$.

 

[Shackleford]

Try it and see what you get.

For TeX/LaTex commands: the command to get $\bar{z}$ is "\ b a r {z}" (no spaces). Also, the command "a^ bc " (no spaces) produces $a^bc$. If you want more than one symbol in a superscript or a subscript, use { }, like this: 'a ^ {b c}" no spaces; that gives $a^{bc}$.

Thanks. I fixed the z-bar. I'm not quite sure that I'm handling everything correctly.

∫ f(z) dz = ∫ f ° φ(t)φ' (t) dt

I get 0, 0, 0, and -1. I think this is somehow related to the fact that it's a closed curve with one loop around the origin and the area of the curve itself.

 

[Ray Vickson]

Thanks. I fixed the z-bar. I'm not quite sure that I'm handling everything correctly.

∫ f(z) dz = ∫ f ° φ(t)φ' (t) dt

I get 0, 0, 0, and -1. I think this is somehow related to the fact that it's a closed curve with one loop around the origin and the area of the curve itself.

If I were doing it I would stay away from parametrization via t, and just deal directly with $\int f(z)\, dz$ as a standard limit like
\lim_{||P|| \to 0} \sum_{z_i \in P} f(z_i) (z_{i+1} - z_i).
Here, $P = P_n = [z_1, z_2,... z_n]$ is a partition of the integration path and $||P||$ denotes something like $\max_i |z_{i+1} - z_i|.$ Basically, $dz = dx + i dy$. If you use a parametrization $z = \varphi(t)$, you need to be careful about keeping the "velocity" constant; otherwise, a $dt$ in one part of the path could give a different $|dz|$ than the same $dt$ in another part of the path. That could lead to things like violations in Cauchy's Theorem, etc. Of course, parametrization would be OK if it were done very carefully, but then it might be more trouble that it is worth.

 

[Shackleford]

If I were doing it I would stay away from parametrization via t, and just deal directly with $\int f(z)\, dz$ as a standard limit like
\lim_{||P|| \to 0} \sum_{z_i \in P} f(z_i) (z_{i+1} - z_i).
Here, $P = P_n = [z_1, z_2,... z_n]$ is a partition of the integration path and $||P||$ denotes something like $\max_i |z_{i+1} - z_i|.$ Basically, $dz = dx + i dy$. If you use a parametrization $z = \varphi(t)$, you need to be careful about keeping the "velocity" constant; otherwise, a $dt$ in one part of the path could give a different $|dz|$ than the same $dt$ in another part of the path. That could lead to things like violations in Cauchy's Theorem, etc. Of course, parametrization would be OK if it were done very carefully, but then it might be more trouble that it is worth.

Ah, that's right. It's easy when the velocity is one.

Oh, I forgot to mention that I found the dt for each parametrization.

So, something like this - (1)(2i) + (1)(-2) + (-1)(-2i) + (1)(2)?

Using both approaches, I get zero. I guess that it means sense because it's a closed curve.