Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 $z=0$? –  anon Mar 13 '12 at 5:17
    
@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. –  Dedede Mar 13 '12 at 5:42

1 Answer 1

up vote 3 down vote accepted

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

share|improve this answer
    
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. –  Dedede Mar 13 '12 at 18:42
    
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). –  Sam Mar 14 '12 at 7:29
    
Many thanks, I satisfied with this. –  Dedede Mar 15 '12 at 0:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.