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This homework problem is giving me some trouble. My current thought process is this:

In order for $ f = u + iv $ to be analytic, it must differentiable. Therefore, it must satisfy the Cauchy-Riemann equation, $ f_{y} = if_{x} $ , and $ f_{x} \text{ and } f_{y} $ must be continuous. The Cauchy-Riemann equation is equivalent to:

$$ u_{x} = v_{y},\\ u_{y} = -v_{x} $$

Because the function must be analytic, they must be differentiable on open neighborhoods of $ z = x + iy $ as well as at z. Because of this fact, I tried doing this:

$$ u_{x} = v_{y} \implies v_{y} = 2x \implies v = 2xy + h(x)\\ u_{y} = -v_{x} \implies v_{x} = 2y \implies v = 2xy + g(y) $$

These two lines imply that $ h(x) = g(y) = constant $, so $ v = 2xy + c $. Thus, functions of the form $ f = x^{2} - y^{2} + i(2xy + c) $ should be analytic. I have two problems. The first problem is that I'm not confident my derivation of v is sound mathematics. Secondly, even if the equation I derived is analytic, how can be certain only equations of this form are analytic? Thanks for the help. I appreciate it.

Max
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    (I suppose your work on an open connected set) It suffices to find one $v_0$ that works. Then the solutions are $v=v_0+C$, $C$ constant. Indeed, if $f=u+iv=u+iv_0$ is analytic, then $v-v_0$ is anaytic and real-valued. This must be constant. – Julien Mar 20 '13 at 12:53
  • I'm not sure I understand fully. Namely this is confusing me: If $ f = u + iv = u +iv_{0} $ is analytic, then $ v - v_{0} $ is analytic and real valued. I'm not sure how this follows. – Max Mar 20 '13 at 13:08
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    $\frac{1}{i}((u+iv)-(u+iv_0))=v-v_0$ is analytic. – Julien Mar 20 '13 at 13:43
  • @Julien Why is an open connected domain required here? – Not Euler Mar 30 '19 at 20:26

2 Answers2

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All harmonic conjugates are all the same up to a constant. (See @Julien's comment for why) so we only need to find a single conjugate. Here's one way to do this question without CR equations

$$u(x,y) = x^2 - y^2 = Re(z^2)$$

This means that it is the real part of an analytic function $z^2 + ci$. Straight away we can say $v$ is the imaginary part of $z^2 + ci$ and so $$v(x,y) = Im(z^2 + ci) = 2xy + c$$

muzzlator
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  • Why $ Re(z^{2}) $? I'm not really sure I follow, sorry. I also don't follow @julien's comment. Sorry to be so dense. – Max Mar 20 '13 at 13:03
  • $z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy)$ and so the real part is $x^2 - y^2$. Julien's comment says that if both $u + v_1 i$ and $u + v_2 i$ are analytic, the difference of analytic functions is analytic and so $(v_1 - v_2)i$ is analytic. Define $u = 0$ and $v = v_1 - v_2$ and suppose $v$ was not constant, then its derivative somewhere won't satisfy the CR equations since $u_x$ and $u_y$ are always $0$. – muzzlator Mar 20 '13 at 13:08
  • Great! That makes perfect sense. Thanks so much. – Max Mar 20 '13 at 13:10
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You started with the CR equations (which any analytic function must satisfy) and deduced $v=2xy + c$. Thus you have proved that if $u=x^2-y^2$ then only functions with $v=2xy+c$ can be analytic. So you're done. (Noticing that $x^2-y^2+i2xy = z^2$, or saying something about real-valued analytic functions being constant, is elegant, but essentially just flair.)

Sean Eberhard
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