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I was working on finding poles of the function

$$f(z) = \frac{1}{1+z^{3/2}}$$

fractional powers should be handled with care, so I'm using the definition via logarithm and for poles we have the following equation (remembering $z = \rho e^{i\theta}$)

$$\begin{align*} e^{3/2\log z} &=-1\\ e^{3/2(\ln \rho + i\theta)} &= e^{i\pi(k+1)}, \ k\in \Bbb Z \end{align*}$$

Where I find, letting $\theta \in (0,2\pi)$ (I'm choosing that branch of the logarithm), poles should be complex numbers such that $$\rho = 1 \\ \theta_k = \frac{2}{3}\pi(k+1), \ k = 0,1,2$$

here i notice that for $k=2$ we get $z_2 = e^{i2\pi} = 1$ so this pole isn't inside this branch cut. The question is should I choose another branch cut in order to include every pole of the function? - in order to answer the original question "where are the poles of $f$?" - or since the branch cut choice is arbitrary The correct answer could also be: " consistently with our branch choice we have two poles located at..."

Thank you for your time, since I'm quite new to complex analysis this kind of function behavior trow me off a little bit but maybe is just me I need to get used to it!

bob
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Turquoise Tilt
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  • Related: https://math.stackexchange.com/questions/5007281/explain-why-the-residue-theorem-seems-to-not-hold-for-frac1zn1/5007325#5007325 – Travis Willse Dec 09 '24 at 09:42

1 Answers1

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It should be $-1={\rm e}^{{\rm i}\pi+2\pi i k}={\rm e}^{{\rm i}\pi(2k+1)}$ for $k\in \mathbb Z$. This gives,$$3{\rm i}\theta_k/2={\rm i}\pi(2k+1)\Rightarrow \theta_k=4\pi k/3+2\pi/3$$ and with the branch cut along the positive real line we only have $k=0$.

bob
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