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I try to show:

Let $f: \mathbb{C} \longrightarrow \mathbb{C} $ be holomorphic with $$\Re(f)+\Im(f)=1 $$ then $ f $ is constant.

($\Re$ = Real Part, $\Im$ = Imaginary Part)

I have certain ideas one is to use Liouvilles theorem about bounded functions but I am not sure if the given information already implies that $f$ is bounded. Consider for example $f(x)=1 + (1 - i) \Re(x)$ then $f$ is not bounded (here it fails because $f$ is not holomorphic). But I don't know how I could use the fact that $f$ is holomorphic to show that its bounded.

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Listing: Just a general remark: questions involving relations between the real and imaginary part cry out lout for applying the Cauchy-Riemann equations. Using Liouville (or Picard...) in this context is not such a good idea, because you won't be able to answer the question using these results if the question is only slightly modified: Let $f: D \to \mathbb{C}$ be holomorphic on a domain (connected open subset of $\mathbb{C}$). Does the conclusion that $f$ is constant still hold? Yes, obviously if you used Cauchy-Riemann. You can use the open mapping theorem but that's a bit more advanced. –  t.b. Jun 11 '11 at 20:34
    
I see, I missed that next time I will be more wise :-) –  Listing Jun 11 '11 at 20:39

6 Answers 6

up vote 4 down vote accepted

$f(z) = u(z) + iv(z), \quad u(z) = 1-v(z)$:

$$\frac{\partial u}{\partial x} = -\frac{\partial v}{\partial x} = \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x}$$

$$\implies \quad 0 = \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$$

q.e.d.

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Hmmm I see no error in this. The best solution I saw so far :-) –  Listing Jun 11 '11 at 21:04
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Hehe, thanks! =) –  Sam Jun 11 '11 at 21:11

This is a special case of a very interesting Liouville-type theorem:

Theorem Let $\Omega$ be an open subset of the complex plane and $f\colon \Omega \to \mathbb{C}$ be holomorphic. If $f(\Omega)$ is contained in a 1-dimensional manifold (i.e. a smooth curve) then $f$ is constant.

The geometric intuition behind this is explained very well in Needham's Visual complex analysis. Locally $f$ acts like a rotation composed with a dilation (amplitwist) so it must be full rank or rank 0, it cannot be rank 1. If the range of $f$ lies in a curve then the rank of $f$ cannot exceed 1 and so it must be 0 everywhere. This means that $f$ must be constant.

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This just seems like a somewhat anemic version of the Open Mapping Theorem. Am I missing some subtlety here? –  Pete L. Clark Jun 11 '11 at 13:08
    
@Pete: I think you're missing nothing, of course: the open mapping theorem contains this one as an immediate corollary. In fact it is true that every holomorphic $f$ whose range has an empty interior is constant. It's just that the enunciate above is easier to prove, because of the additional information on $f(\Omega)$ (its tangent space must be of dimension 0 or 1). I find it instructive. –  Giuseppe Negro Jun 11 '11 at 16:27
    
Thank you for the answer, but why exactly is $f$ contained in a 1-dimensional manifold? –  Listing Jun 11 '11 at 16:38
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@Listing: The constraint $\rm{Re}(f)+\rm{Im}(f)=1$ forces the range of $f$ into the line $\{x+i y \mid x+y=1\}$. –  Giuseppe Negro Jun 11 '11 at 16:50
    
Ah now I see it, thank you for enlighting me. I will try to find a down-to earth proof for this theorem. –  Listing Jun 11 '11 at 16:51

Hint: use Cauchy-Riemann equations.

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The image of the plane under this map must line inside of a line. Apply the open mapping theorem: nonconstant analytic functions are open maps. Since the image of any open set cannot be open, this function must be constant.

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The condition is equivalent to $g(z)=(1-i)f(z)$ is real for all complex $z$. Note that if $f=k+ik$, then $(1-i)k(1+i)$ is real. The imaginary part of $g(z)$ vanishes - it's an equivalent way to phrase your condition.

The derivatives of $g(z)$ with respect to $z$, which exists because $g(z)$ is holomorphic just like $f(z)$, can be calculated as $$\frac{dg(z)}{dz} = \frac{dg(z)}{dz_1} = \frac{\partial g_1(z)}{\partial z_1} + i \frac{\partial g_2(z)}{\partial z_1} = \frac{\partial g_2(z)}{\partial z_2} + i\frac{\partial g_2(z)}{\partial z_1}$$ by the Cauchy-Riemann equations. I used one of the Cauchy-Riemann equation to rewrite the partial derivative of the real part (the first term) to a derivative of the imaginary part. Note that the first step was to define the derivative with respect to $z$ as the derivative with respect to the real part of $z$. It's really the same thing for holomorphic functions.

If the first or second term has a wrong sign, or if the $i$ should be in the first term and not the second, it doesn't make a difference. But because the imaginary part of $g$, $g_2$, was assumed to be zero, $dg_2(z)/dz$ is zero as well, and therefore $g$ is constant, and therefore $f$ is constant, too.

A more general message: Pretty much any similar constraint unnaturally depending both on the real and imaginary part - instead of the whole $z$ itself - will force any holomorphic function to be constant. Holomorphic functions are really meant to make the operations "real part" and "imaginary part" inappropriate. That's because the "real part" and "imaginary part" are not holomorphic functions themselves.

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Pleasure. Maths is trivial - once it's proved. ;-) –  Luboš Motl Jun 11 '11 at 11:41
    
Hmm you say $g(z)$ is real but I am sceptical. Let $f(z)=a+bi$ then $a+b=1$ and you say that $(1-i)(a+b i)=a+b - i (a - b)=1- i(a-b) \in \mathbb{R}$ But that is equivalent to $a=b$ but as $a=1-b$ it follows that $b=1-b$ and therefore $b=1/2$, which is obviously not always true?? –  Listing Jun 11 '11 at 16:23

We can use Picard's Theorem in this one too. Our function is entire and $0,2$ are not in the image of $f$. This forces $f$ to be constant.

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To paraphrase a comment by John Klein: [Beni], [y]ou're using an anti-tank missile to prove a result that requires no more an a pea-shooter. In other words: Do you prove every result in complex analysis with the conclusion that a function is constant using Picard? :) –  t.b. Jun 11 '11 at 15:12
    
Yes, I was thinking that too... :) I think I've given at least 5 proofs using Picard, to relatively easy questions. I'll try and stop myself from doing that in the future. :) –  Beni Bogosel Jun 11 '11 at 15:20

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