Conservativeness of $e^z$

Question

So I recently have started trying to dive deeper into the realm of complex analysis, but to get there I needed some notions of multivariable/vector calculus for stuff like line integrals, etc... I don't own a book or anything to follow, and instead I'm just using the internet for stuff (after all this is only for curiosity anyways), though I do have good background knowledge on single-variable calculus, real analysis, and even some complex analysis stuff (simple things like euler's formula and the genre).

Generally there are a lot of resources to see and things are going great, but while down the rabbit-hole of path independence (as a general concept), I got introduced to the concept of conservativeness. Now although I don't intend going deep into those branches of math quite yet, it left me with a question. If I understand it correctly, holomorphic functions like $e^z$ on a simply connected region are path-independent, and have contour integrals equal to 0. But if we treat $e^z$ as a vector field $(x,y) \rightarrow (e^x \cos(y), e^x \sin(y))$, the curl not only is different from $0$, but it makes intuitive sense since there are many "circles" in the field due to how the involved cos and sin manipulate angles.

So my question is, did I get something wrong, and if not, why isn't the result of a contour integral the same as the result of the equivalent vector integral with my defined vector field?

Answer 1

While a function from $\mathbb{C}$ to $\mathbb{C}$ can naturally be considered a vector field, its contour integral in complex analysis means something different from the contour integral of the vector field. If $\vec F(x,y) = (e^x \cos y, e^x \sin y)$ and $\gamma$ is a curve (whether closed or not) in $\mathbb{C}$ then usually $$ \int_\gamma e^z dz \ne \int_\gamma \vec F \cdot d\vec r $$ Actually, this should be clear from the get-go before we do any computations: The LHS is a complex number in general, while the RHS is real.

What's going on? Note that $dz = dx + i dy$, while $d\vec r = (dx,dy)$. Multiplication in $\mathbb{C}$ is not the same as a dot product, which is why the two integrals are different.

Expanding out the complex integral using $dz = dx + i dy$, we can see that the real and imaginary parts of $\int_\gamma e^z dz$ are contour integrals of certain vector fields, but they are not the contour integrals of $\vec F$. Observe: $$ \int_\gamma e^z dz = \int_\gamma e^z (dx + i dy) = \int_\gamma (e^x \cos y + i e^x \sin y)(dx + i dy) $$ Multiplying it out, you get \begin{eqnarray} \int_\gamma e^z dz &=& \int_\gamma (e^x\cos y dx - e^x\sin y dy) + i (e^x\cos y dy + e^x \sin y dx)\\ &=& \left(\int_\gamma e^x \cos y dx - e^x \sin y dy\right) + i\left( \int_\gamma e^x\sin y dx + e^x\cos y dy\right)\\ &=& \int_\gamma (e^x\cos y, -e^x\sin y) \cdot d\vec r + i \int_\gamma (e^x\sin y, e^x \cos y)\cdot d\vec r \end{eqnarray} So the real and imaginary parts of $\int_\gamma e^z dz$ are integrals of different vector fields than $(e^x \cos y, e^x \sin y)$. Note that $(e^x\cos y, -e^x\sin y)$ and $(e^x\sin y, e^x \cos y)$ are conservative vector fields.

In general, if $f(z) = f_1(z)+if_2(z)$, you'll get $$ \int_\gamma f(z) dz = \int_\gamma (f_1,-f_2)\cdot d\vec r + i \int_\gamma (f_2,f_1)\cdot d\vec r $$ The Cauchy-Riemann equations ensure that those two vector fields are irrotational whenever $f$ is holomorphic.

Hope this helps!

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Addendum: You might wonder what is the complex analysis interpretation of $\int_\gamma \vec F \cdot d\vec r$. In that integral, we are integrating $f_1 dx + f_2 dy$, which would be the real part of $f(z) (dx - i dy)$. Observe $$ f(z)(dx-idy) = (f_1 + i f_2) (dx - idy) = (f_1 dx+f_2dy) + i (f_2dx - f_1dy) $$ Hence, if $f(z) = f_1(z)+i f_2(z)$ and $\vec F = (f_1,f_2)$, we have $$ \int_\gamma f(z)d\bar z = \int_\gamma \vec F \cdot d\vec r + i\int_\gamma (f_2,-f_1)\cdot d\vec r $$ Note that $(f_2,-f_1)$ is simply $\vec F$ rotated $90^\circ$ counterclockwise.

Answer 2

Let $f=u+iv$. Then $$ \int_\gamma f dz=\int_\gamma (u+iv)(dx+idy)=\int_\gamma udx-vdy+i\int_\gamma vdx+udy. $$ Now apply what you know about conservative fields to $(u,-v)$ and $(v,u)$ in your specific example.