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The infinite Taylor series expansion of $\sqrt{1+is}$ around $s=0$ is $$\sqrt{1+is}=\frac{1}{\sqrt{\pi}}\sum_{n=0}^{\infty} (-i)^n \frac{\Gamma(n+\frac{1}{2})}{n!}s^n$$ where $i$ is the imaginary unit. I am trying to find the infinite Taylor series expansion around $s=0$ of $$\frac{1}{\sqrt{1+is}-\left(\frac{1}{2}+\frac{i}{4}\right)}$$ I am having trouble with this problem, any ideas?

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  • $\begingroup$ Why choose a title which does not describe what the question is asking? $\endgroup$ – Did Feb 23 '17 at 6:23
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Complement by the conjugate: $$f(s)\dfrac{\sqrt{1+is}+\frac{2+i}{4}}{(\sqrt{1+is}+\frac{2+i}{4})(\sqrt{1+is}-\frac{2+i}{4})} = \dfrac{\sqrt{1+is}+\frac{2+i}{4}}{\frac{13-4i}{16}+is}.$$ Now for the denominator use, the Taylor expansion of $\dfrac{1}{a-bz} = \dfrac{1}{b}\dfrac{1}{\frac{a}{b}-z}$. The first part of the fraction, $\dfrac{\sqrt{1+is}}{\frac{13-4i}{16}+is}$ is to be computed by multiplying the two corresponding Taylor expansions. The second part $\dfrac{\frac{2+i}{4}}{\frac{13-4i}{16}+s}$ is readily computed by the Taylor expansion of $\dfrac{1}{a-bz} = \dfrac{1}{b}\dfrac{1}{\frac{a}{b}-z}$.

This will avoid the taking derivative approach and will give you everything in terms of the general sum notations, but the final answer will undoubtedly be somewhat unpleasant.

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  • $\begingroup$ I'm not sure if your $f(s)$ is correct. Could you show the steps of how you obtained $f(s)$. $\endgroup$ – Hogg Feb 22 '17 at 18:05
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    $\begingroup$ edited just now. $\endgroup$ – dezdichado Feb 22 '17 at 18:29

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