Let $V$ be an open connected subset of $\mathbb{C}$.
Let $f:V\rightarrow\mathbb{C}$ be a function whose derivative is $0$ on $V$.
How do I prove that $f$ is a constant on $V$?
I know that $V$ is path-connected, but I don't know whether this helps.
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Can you see the proof for a real-valued function on a real interval $(a,b)$, that if $f'(x)=0$ at every point in that interval, then $f$ is constant there? – hardmath Apr 13 '14 at 03:47
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@hardmath Yes sure. I know an argument by Fermat for that one.. – user140374 Apr 13 '14 at 03:57
4 Answers
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Use the Cauchy-Riemann equations:
$$ f(x + iy) = u(x, y) + iv(x, y) $$
Where $u$ and $v$ are real valued functions. We have the constraints:
$$ \frac{\partial u}{\partial x} = \frac{\partial v }{\partial y} \\ \frac{\partial u}{\partial y} = -\frac{\partial v }{\partial x} $$
Giving:
$$ f'(x + iy) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} $$
If $f' = 0$ then we have that $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 0$. This then entails that $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial y} = 0$. If $\frac{\partial u}{\partial x} = 0$ then $u = \alpha(y)$. If you do all of them then you find that $u = \alpha(y) = \beta(x)$ and $v = \gamma(y) = \varphi(x)$. The only way that $\alpha(y) = \beta(x)$ for all values of $x$ and $y$ is if $\alpha$ and $\beta$ have no dependence on either $x$ or $y$ and thus are constants. The same follows for $v$. Therefore $u = A$ and $v = B$ and $f(z) = A + Bi$.
Jared
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@user140374 The $\alpha$, $\beta$, $\gamma$, and $\varphi$ are just some functions of single variables. Since the partial is $0$, the original multi-variable function must actually have been a function of a single variable. They must be equal so that $\alpha(y) = \beta(x)$. Let's choose a value of $x = x_0$ and $y = y_0$, then we must have $\alpha(y_0) = \beta(x_0)$. What if we change $y$? Then $\alpha(y_0 + h) \neq \alpha(y_0)$ unless $\alpha(y_0 + h) = \alpha(y_0) = A$, some constant. But if they aren't equal, then $\alpha(y_0 + h) \neq \beta(x_0)$, a contradiction. – Jared Apr 13 '14 at 03:54
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@user140374 If the domain is not connected, then you could have several distinct values for $f$ on each disconnected region. It would basically be $f(z) = a_n$ when $z \in D_n$ and $\forall m, n: n\neq m \leftrightarrow D_n \cap D_m = \emptyset$. This would mean that $f$ is not exactly constant (although it should be a "constant step function"). – Jared Apr 13 '14 at 04:07
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Consider a point $z_0 \in V$. Now consider any other point $w \in V$. Since $V$ is connected, it is possible to find a curve starting at $z_0$ and ending at $w$. Let's call this curve $C_w$. We then have $$f(w) - f(z_0) = \int_{C_w} f'(z) dz = \int_{C_w} 0 \cdot dz = 0$$ Hence, we have $f(w) = f(z_0)$. This is true for any $w \in V$. Hence, $f(w) = f(z_0)$ and hence $f(w)$ is a constant in $V$.
user141421
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wait but, to clarify, you're assuming path connected instead of connected? – BCLC Oct 23 '21 at 17:10
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wait never mind i remember now: open and connected subset of Rn implies path connected: https://math.stackexchange.com/questions/1088685/showing-an-open-connected-subset-of-mathbbrn-is-path-connected – BCLC Oct 23 '21 at 17:18
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Do we need simply connected domain instead of open, connected domain for the above integral to be well-defined? – Sourav Ghosh Feb 18 '22 at 14:12
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Hint: Choose a path $\gamma$ connecting two given points in $V$, and evaluate
$$\int_{\gamma} f'(z) dz$$
two different ways.
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I haven't learned line integral though.. Isn't there a way to prove this not using line integral? – user140374 Apr 13 '14 at 03:39
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@user140374: Have you learned the regular multivariable line integral? You can just treat the function as a pair of real-valued functions with two real-valued inputs. – user2357112 Apr 13 '14 at 03:42
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@user140374 In that case, what have you learned? (Please note that adding additional details to the original question when asking will help a lot in getting answers that are actually relevant to you.) – Apr 13 '14 at 03:43
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@user140374: Huh. You're doing complex analysis before multivariable calculus? Weird. – user2357112 Apr 13 '14 at 03:44
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@user2357112 I did multivariable calculus long time ago but not precisely. The only precise part i have done in multivariable calculus is just inverse,implicit,rank theorem. – user140374 Apr 13 '14 at 04:03
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@T.Bongers I'm trying to understand your answer. Is it true that "For any distinct point $x,y$ in an open connected subset $V$, there exists a differentiable function $f:[0,1]\rightarrow V$ such that $f(0)=x,f(1)=y$? – user140374 Apr 13 '14 at 04:04
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@user140374 Yes, there is; but really all that's necessary for the path integrals to make sense is piecewise smooth. – Apr 13 '14 at 04:59
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You can connect any point to a point $z_0$ by a polygonal line. Along each segment you can write $f(z) = f(z_i + t (z_f - z_i))$, and write this as two real functions of the real variable $t$, one for the real part and one for the imaginary part. Write the derivative in terms of them, you see both components are zero along the segment, so both components are constant and the function itself is constant on that segment, and thus over each segment of the path, and so over the full region.
vonbrand
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I succeed in proving the theorem in this way. Thank you. BTW, how do you define polygonal line precisely? I rather showed that given $\alpha,\beta$, there exist finite points $\alpha=z_0,\cdots,z_n=\beta$ such that $z_i-z_{i+1}$ is either purely imaginary or purely real, and the line segment connecting these two are in $V$, then apply Cauchy-Riemann equation. This formulation is precise but really messy when written in meta-language. – user140374 Apr 13 '14 at 06:02
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I know it is a mess to write up precisely, but it is easy to explain. BTW, straight lines between $z_i$ to $z_f$ are just $z=z_i + (z_f -z_i) t$ with real $t$ between 0 and 1. And my definition of a connected region if based on polygonals connecting any pairof points – vonbrand Apr 13 '14 at 12:39