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I'm working on an exercise which asks to apply the representation $f(z)=w_0+\zeta(z)^n$ to $\cos z$ with $z_0=0$, and determine $\zeta(z)$ explicitly. (This is from Ahlfors, page 133, ex. 3 by the way).

Now $\cos(0)=1$, so $w_0=1$. Also, $\cos z-1$ has a zero at $z_0=0$ of order $2$. So I can write $\cos z-1=z^2g(z)$ with $g(z)$ analytic at $z_0$, and $g(z_0)\neq 0$. Taking $z^2g(z)=\zeta(z)^2$, I just find $\zeta(z)=\sqrt{\cos z-1}$, so $$ \cos z=1+\sqrt{\cos z-1}^2. $$

I'm worried, have I done this correctly? I don't feel like I've done anything of substance, so I think the exercise might have been asking for something else.

I think Ahlfors is referring to a theorem stating

Suppose that $f(z)$ is analytic at $z_0$, $f(z_0)=w_0$ and that $f(z)-w_0$ has a zero of order $n$ at $z_0$. If $\epsilon>0$ is sufficiently small, there exists a corresponding $\delta>0$ such that for all $a$ with $|a-w_0|<\delta$ the equation $f(z)=a$ has exactly $n$ roots in the disk $|z-z_0|<\epsilon$.

Later discussion states that under the assumption of this theorem, we can write $$ f(z)-w_0=(z-z_0)^ng(z) $$ where $g(z)$ is analytic at $z_0$ and $g(z_0)\neq 0$. Choose $\epsilon>0$ so that $|g(z)-g(z_0)|<|g(z_0)|$ for $|z-z_0|<\epsilon$. In this nbhd is's possible to define a single-valued analytic branch of $\sqrt[n]{g(z)}$ which we denote by $h(z)$. We have thus $$ f(z)-w_0=\zeta(z)^n,\qquad \zeta(z)=(z-z_0)h(z). $$ Since $\zeta'(z_0)=h(z_0)\neq 0$, the mapping $\zeta\zeta(z)$ is topological in a neighborhood of $z_0$.

I think the implied representation is related to the conventions here.

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  • $\begingroup$ So this representation you speak of... you mean you want to write $\cos$ as a sum of a constant and a power of a function with a zero at $z=0$? $\endgroup$
    – anon
    Mar 13, 2012 at 5:17
  • $\begingroup$ @anon That's the heart of my problem, I don't know exactly what representation Ahlfors' means. I've tried to add relevant discussion from the section where I find this problem. $\endgroup$
    – Dedede
    Mar 13, 2012 at 5:42

1 Answer 1

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HINT: Note that $1-\cos(2z)= 2\, \sin^2(z)$.

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  • $\begingroup$ Thanks Sam. So this implies $\sqrt{\cos z-1}=i\sqrt{2}|\sin(z/2)|$? Is that all there is to it? I'm just not sure of the final form for $\zeta(z)$ I'm supposed to have. $\endgroup$
    – Dedede
    Mar 13, 2012 at 18:42
  • $\begingroup$ Note that the square root of $\cos z - 1$ is a holomorphic function, so you can't have absolute values around $\sin$. If you insist on the form $\cos z = 1 + z^2 \zeta(z)^2$, then $\zeta(z) = i\sqrt{2} \dfrac{\sin(z/2)}z$ would be a possibility by the above identity (the other alternative being the same with a minus sign). $\endgroup$
    – Sam
    Mar 14, 2012 at 7:29
  • $\begingroup$ Many thanks, I satisfied with this. $\endgroup$
    – Dedede
    Mar 15, 2012 at 0:30

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