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Verify that each given function $u$ is harmonic (in the region where it is defined) and then find a harmonic conjugate of $u$.

(a) $u=y$

I was able to verify that $u$ is harmonic pretty easily. But I found that the harmonic conjugate of $u$ is $-x+a$. Thus, I get that $f(z)= y + i(-x+a)$. Using that $y = \frac{z-\bar{z}}{2i}$ and that $x=\frac{z+\bar{z}}{2}$, I get that $f(z) = -i(\bar{z}+a)$, which as I understand is not an analytic function. I would like to know if $f(z)$ is indeed analytic, and whether or not having a harmonic function $u$ along with its conjugate $v$ in $f(z)$ guarantees having an analytic function $f(z)$?

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    $\begingroup$ I think there's an error in your computation. In your example, $f(z) = -iz + ia$, which is affine and hence analytic. $\endgroup$ – Travis Willse Feb 10 '19 at 21:45
  • $\begingroup$ In any case, one can always find a harmonic conjugate locally but not always globally, at least on domains that are not simply connected. The standard example is $z \mapsto \log |z|$ on $\Bbb C - {0}$. $\endgroup$ – Travis Willse Feb 10 '19 at 21:52
  • $\begingroup$ I was wondering what is meant by locally? Taking an example from the harmonic function $u=y$, does that mean that the domain of $v$ (harmonic conjugate) can be found in some neighborhood of the domain of $u$? $\endgroup$ – K.M Feb 10 '19 at 21:58
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    $\begingroup$ On a domain that is not simply connected, one is actually guaranteed more than local existence but one is not guaranteed existence of a harmonic function on all of its domain. (Here, local existence would just mean that for any point $z_0$ in the domain of $u$ that there is a neighborhood $A$ of $z_0$ and a harmonic conjugate $v$ of $u\vert_A$ on $A$.) I've written up an answer that explains more precisely what is guaranteed. $\endgroup$ – Travis Willse Feb 11 '19 at 4:03
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Suppose $D \subset \Bbb C$ is a domain and $u : D \to \Bbb R$ is a harmonic function.

If $D$ is simply connected, then $u$ admits a harmonic conjugate (globally). Explicitly, up to an overall constant it is (for any choice of base point $z_0 \in D$) $$v(z) := \int_{z_0}^z -u_y \,dx + u_x \,dy .$$ Computing the exterior derivative of the integrated $1$-form gives $$d(-u_y \,dx + u_x \,dy) = (u_{xx} + u_{yy}) \, dx \wedge dy = 0 ,$$ where the second equality just uses that $u$ is harmonic, so $-u_y \,dx + u_x \,dy$ is closed and hence (since $D$ is simply connected) exact, and thus the integral is well-defined. If $-u_y \,dx + u_x \,dy = df$, then $v(z) = \int_{z_0}^z df = f(z) - f(z_0)$.

Now, differentiating shows that the pair $(u, v)$ satisfies the Cauchy-Riemann equations---in fact the definition of $v$ is rigged precisely so that this happens---or equivalently that the function $f : D \to \Bbb C$, $f := u + i v$, is analytic on $D$.

On the other hand, if $D$ is not simply connected, $u$ may not admit a global harmonic conjugate, i.e., one on all of $D$, whereas the above argument shows that for any simply connected subset $A \subset D$, $u\vert_A$ does admit a harmonic conjugate $v : A \to \Bbb R$. To find a counterexample, inspecting the argument shows that we certainly need a $1$-form that is closed but not exact (any such $1$-form must have domain that is not simply connected). The standard example is $$\frac{-y \,dx + x \,dy}{x^2 + y^2}.$$ On any simply connected subset of $\Bbb C - \{ 0 \}$ this is the exterior derivative of some function, and the Cauchy-Riemann equations imply that (the negative of) its harmonic conjugate has exterior derivative $$\frac{x \,dx + y \,dy}{x^2 + y^2} .$$ But this expression defines a $1$-form on all of $\Bbb C - \{ 0 \}$, and this $1$-form is exact: An easy integration shows that it's just the exterior derivative of $$u(z) := \frac{1}{2} \log (x^2 + y^2) = \log |z| .$$ By construction, $u$ does not have a harmonic conjugate on all of $\Bbb C - \{ 0 \}$. Up to additive constants, the harmonic conjugates of the restrictions of $u$ to simply connected subsets are branches of the argument function.

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