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Find the multiplicity of the root at $z_0$ for these functions

i) $$ \begin{align} f(z)= e^{zcos(z)-z}-1, z_0=0 \end{align} $$

Let $z_0$ be a root of a holomorphic function $f$ , and let n be the least positive integer such that, the nth derivative of $f$ evaluated at $z_0$ differs from zero.

Based on the definition , I evaluated the derivatives and then took their value at $0$

$f'(z)=e^{z (\cos (z)-1)} (-z \sin (z)+\cos (z)-1)$ and $f'(0)=0$

$f''(z)=e^{z (\cos (z)-1)} \left(z^2 \sin ^2(z)+2 (z-1) \sin (z)+\cos ^2(z)-(z+2 z \sin (z)+2) \cos (z)+1\right)$ and $f''(0)=0$

$f'''(z)=\\e^{z (\cos (z)-1)} \left(z \sin (z)-3 \cos (z)+(-z \sin (z)+\cos (z)-1)^3+3 (2 \sin (z)+z \cos (z)) (z \sin (z)-\cos (z)+1)\right)$

and $f'''(0)=-3$ therefore the multiplicity of the root at$z_0=0$ is 3.

My question is , is there another more efficient way to determine the order of the root (or the pole maybe?) instead of calculating the derivatives until we get a non zero value? I haven't found any solved examples and I don't know if there is a diferrent standard method of determining this.

helplessKirk
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1 Answers1

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Often, it is easier/more efficient to use the Taylor expansions of the involved functions, when one looks at compositions. Here, we have

$$z\cos z - z = z\left(1 - \tfrac{z^2}{2} + \tfrac{z^4}{24} - \dotsc \right) - z = - \tfrac{z^3}{2} + O(z^5),$$

and inserting that into the Taylor expansion of $e^w$, we find

$$e^{z\cos z - z} = 1 - \frac{z^3}{2} + O(z^5),$$

from which it is immediate that $e^{z\cos z-z}-1$ has a zero of order $3$ at $0$.

Daniel Fischer
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  • Thanks for your answer! Espesially in this case using the Taylor expansions was indeed much simpler. – helplessKirk Aug 24 '14 at 12:55
  • sorry if I'm asking something stupid but , can you tell me why inserting the $-\frac{z^3}{2}+ O(z^5)$ gives us $1 - \frac{z^3}{2} + O(z^5)$ as a result? – helplessKirk Aug 24 '14 at 15:13
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    Let's write it as $w(z) = -\frac{z^3}{2} + R(z)$. Then $$e^{w(z)} = 1 + w(z) + \frac{w(z)^2}{2} + \frac{w(z)^3}{6} + \dotsc,$$ and the factors $w(z)^k$ for $k\geqslant 2$ all are divisible by $z^6$, so we have $e^{w(z)} = 1 + w(z) + O(z^6)$. Now expanding $w(z)$ again, we have $1 -\frac{z^3}{2} + O(z^5) + O(z^6)$, and since $z^6$ vanishes faster than $z^5$, we can absorb that into a new $O(z^5)$ term. – Daniel Fischer Aug 24 '14 at 15:21
  • Thank you very much, it's all clear now. – helplessKirk Aug 24 '14 at 15:26