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I am reading Donald Sarason's "Notes on Complex Function Theory". I have two questions about the following (taken from page $88$ of the book):

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  1. Why did we had to use $g$ ? We already had $f$ which was claimed to be holomorphic, so it seems that$-\frac{\partial u}{\partial y}$ is the harmonic conjugate of $u$

  2. I'm guessing I am wrong in my statement that $-\frac{\partial u}{\partial y}$ is the harmonic conjugate of $u$, what is the harmonic conjugate ?

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    $\begingroup$ Look at the definition of $f$ again. $-\partial u/\partial y$ is the harmonic conjugate of $\partial u/\partial x$, not $u$. $\endgroup$ – Potato Jun 27 '13 at 22:11
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    $\begingroup$ If $u$ is a harmonic function then there is a holomorphic function of the form $u+iv$, where $v$ is unique up to a constant. We call $v$ the harmonic conjugate. Why would we think $-\frac{\partial u}{\partial y}$ is a harmonic conjugate of $u$, i.e. why believe $u-i\frac{\partial u}{\partial y}$ is a holomorphic function of $x+iy$? $\endgroup$ – anon Jun 27 '13 at 23:03
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Why did we had to use $g$

Because we want to prove the theorem stated above. The existence of $g$ with $\operatorname{Re}g=u$ is a part of the theorem.

it seems that $−\dfrac{\partial u}{\partial y}$ is the harmonic conjugate of $u$

This is incorrect, as Potato and anon already pointed out.

what is the harmonic conjugate of $u$?

Since $\operatorname{Re}g=u$, the function $\operatorname{Im}g$ is a harmonic conjugate of $u$. (Not the harmonic conjugate, since we can add any constant and get other conjugate functions.)

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  • $\begingroup$ Thanks for the answer! Can I conclude that $u+c$ is $Im(g)$ (which is a harmonic conjugate) ? (where $c$ is a constant) $\endgroup$ – Belgi Jun 28 '13 at 18:26
  • $\begingroup$ @Belgi What would be the basis for such conclusion? Is there a theorem that the imaginary part of a holomorphic function is equal to its real part, plus a constant? $\endgroup$ – ˈjuː.zɚ79365 Jun 28 '13 at 23:47
  • $\begingroup$ No. My reasoning is this: We had $\frac{\partial u_{1}}{\partial x}=\frac{\partial u}{\partial x}$ and we concluded $u-u_{1}=c$ for some constant $c$, but since we assumed $g(z_{0})=u(z_{0})$ we got that $u=u_{1}$. So we got that the real part of $g$ is $u$. Now I'm looking at the imaginary part, we got: $\frac{\partial u}{\partial y}=\frac{\partial u_{1}}{\partial y}$, doesn't this equality means that $u$ {[}lus a constant is the imaginary part of $g$ ? $\endgroup$ – Belgi Jun 29 '13 at 7:10
  • $\begingroup$ @Belgi Are you misreading $v_1$ as $u_1$, by any chance? The imaginary part of $g$ is denoted $v_1$ in the proof. $\endgroup$ – ˈjuː.zɚ79365 Jun 29 '13 at 7:12
  • $\begingroup$ But they used C-R equations to express the imaginary part of $g$ using $u_1$, in the first line that have $g'=..=..$ $\endgroup$ – Belgi Jun 29 '13 at 7:55

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