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Let $f(z)=u(x,y)+ i v(x,y)$ be a holomorphic function. I often find it difficult to deduce the form of $f$ as a function of $z=x+i y$ only (for example: $f=\sin x \cosh y + i \cos x \sinh y$ cas be written as $f(z)=\sin z$).

However, my book states that given $u$ and $v$ or even $u$ or $v$ only we can deduce the $z$-dependance of $f$ using the following relations:

$f(z)=u(z,0)+i v(z,0)$

$f(z)= 2 u (z/2, -i z/2)$

$f(z)= 2 i v(z/2, -iz/2)$

Can anyone help me understand how they come about?

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Hmm. I tried to answer this, but I'm now confused. Take $f(z) = \sin z + 1$. Then $u(x,y) = \sin x \cosh y + 1$, and $v(x,y) = \cos x \sinh y$, unless I'm very much mistaken. But then $f(0) = 1 \not= 2u(0,0)$ as the second equation suggests it should. Is there some condition you're missing here? – Ben Millwood Jun 10 '12 at 12:19
I think it works. $2u(0,0)=2 \sin(0) \cosh(0) + 1 = 1$ as $\sin(0)=0$ – malina Jun 10 '12 at 12:59
No, $2u(0,0)=2(\sin 0\cosh 0 + 1) = 2$. – Ben Millwood Jun 10 '12 at 14:08
Okay, simpler example: $f(z) = 1$. Then $u(x,y) = 1$ and $v(x,y) = 0$, and $2u(z/2,-iz/2) = 2\not=f(z)$. – Ben Millwood Jun 10 '12 at 14:09
You're absolutely right, thanks :) – malina Jun 10 '12 at 15:17
up vote 2 down vote accepted

For a start, it's worth mentioning that you kind of already have $f$ as a function of $z$ only, just using the real part and imaginary part functions a lot. So it's a slightly non-obvious question what we're really trying to achieve here – basically, we want a "neat" expression for $f$, but that's not a terribly mathematical concept. With that in mind, I'll go ahead and see if I can work out what the relations mean.

  • $f(z) = u(z,0) + iv(z,0)$: This is a bit of a cheat. A priori, $u$ and $v$ need not be capable of taking a complex argument at all. But if they do extend to complex-differentiable functions on $\mathbb C$, then they extend to a function that agrees with $f$ on the real line (because on the real line, the equation is completely straightforward), and it is a general theorem that if two complex-differentiable functions agree on any non-discrete set then they must be equal (as a consequence of the fact that the zeroes of a nonzero complex differentiable function are isolated, so if $f$ and $g$ agree on non-isolated points, then $f-g$ is zero on non-isolated points, so $f - g = 0$, i.e. $f = g$).
  • The latter two equations do not hold. For example, take $f(z)=1$, so that $u(x,y)=1$ and $v(x,y)=0$. Then $f(0)=1$, but $2u(0,0)=2$ and $2v(0,0)=0$.
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