What is the residue of $\frac{z}{Log{z}}$ at $z=1$? I tried expanding log into it's series, factoring out a $z-1$ and then applying the geometric series expansion, but I'm just getting a series that starts at $k=1$....
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Note that the residue doesn't change under translations i.e. $z\to z+z_0$; however, the pole itself moves by $-z_0$. Try shifting the pole to $0$. – Semiclassical Aug 11 '14 at 02:28
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Note that, as $\frac d{dz}\log z = \frac1z$, which $\neq 0$ at $1$, the pole is a simple one; so the residue is $\frac{z}{\frac 1z} = z^2$ at 1, or $1$.
Clinton Bradford
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3@Anthony : There is no such thing as "being allowed" to do that! There is only reasoning and proof. =) – Patrick Da Silva Aug 11 '14 at 02:26
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1Not sure who's forbidding you; the residue at a simple pole is $\lim_{z \to a} \frac{f(z)}{g(z)}{(z-a)} = \lim f(z) \frac{1}{\frac{g(z)-g(a)}{z-a}} = \frac {f(a)}{g'(a)}$ – Clinton Bradford Aug 11 '14 at 02:27
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Another method:
$$\frac{z}{\log(z)} = \frac{z}{\log(1+(z-1))} = \frac{z}{(z-1) - (z-1)^2/2 + ...}$$ and so $\lim z \rightarrow 1 \frac{z(z-1)}{\log(z)} = \lim \frac{z}{1-(z-1)/2 + ...} = 1$ as we found above.
Now, suppose we want the residue for $H(z) = \frac{f(z)}{g(z)}$ at $z=a$ where the pole is simple. Then write $g(z) = g(a) + g'(a)\cdot (z-a) + g''(a)\cdot (z-a)^2/2 + ...$. Then, $\lim_{z\rightarrow a} \frac{f(z)\cdot (z-a)}{g(a) + g'(a)\cdot (z-a) + O(2)} = \lim \frac{f(z)}{g'(a) + (z-a)*(?)} = \frac{f(a)}{g'(a)}$ since $g(a) = 0$. In fact, we can generalise such that for an $n$-pole (where $f$ and $g$ are both holomorphic), the residue is given as $\frac{nf^{(n-1)}(a)}{g^{(n)}(a)}$.
Christopher K
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