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For the complex-valued function

$$f(z) = \left(\frac{\sin 3z}{z^2}-\frac{3}{z}\right)$$

classify the singularity at $z=0$ and calculate its residue.

Attempt at Solution

Rewriting $f(z) = \left(\frac{\sin (3z) - 3z}{z^2}\right)$, I'm not sure whether the singularity at 0 is removable or a pole because although both numerator and denominator vanish at $z=0$, the sine function is involved and the degree in the denominator is $2$.

Assuming it's a double pole at $z=0$, I calculated the residue to be $0$.

Comments & clarifications welcome. Thank you.

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Hint: What is the behavior of $\sin(x)/x$ near zero? Now, what is the behavior of $[\sin(x)-x]/x$ near zero? –  Jonathan Christensen Dec 16 '12 at 0:02
    
You can consider the Laurent series of f at z=0. –  Ben Dec 16 '12 at 0:04
    
@Jonathan - yes, I can see your logic in the case where $x$ is a real variable. But for complex variables, is the behaviour of $\sin z$ near 0 (for e.g. along the imaginary axis) as predictable? –  Conan Wong Dec 16 '12 at 5:15
    
@Ben - thanks, yes, makes sense. –  Conan Wong Dec 16 '12 at 5:19
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3 Answers

up vote 2 down vote accepted

Another way:

$$\lim_{z\to 0}\left(\frac{\sin 3z}{z^2}-\frac{3}{z}\right)=\lim_{z\to 0}\frac{\sin 3z-3z}{z^2}\stackrel{\text{L'Hospital}}=\lim_{z\to 0}\frac{3\cos 3z-3}{2z}\stackrel{\text{L'H}}=\lim_{z\to 0}\frac{-9\sin 3z}{2}=0$$

So the singularity is a removable one.

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DonAntonio, thanks. Any extra care needed when applying L'Hopital's Rule for complex variables? Or is it as straightforward as in the real variable case? –  Conan Wong Dec 16 '12 at 5:17
    
In this case it is basically the same as in the real case. In the $\,\frac{\infty}{\infty},$ case it also works...*but*...some care must be put in case we have that not just the variable $\,z\to \infty\,$ but only $\,Re(z)\to\infty\,\,\,or\,\,\,Im(z)\to\infty\,$ , as this can have some rather funny effects on some particular functions –  DonAntonio Dec 16 '12 at 10:35
    
Thank you very much for the clarification. –  Conan Wong Dec 17 '12 at 16:03
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The easiest thing in this cases (for me) is just to calculate the principal part of the Laurent expansion at zero.

$\sin (3z) = 3z-9z^3/2+...$ so $f(z)= 3/z-9z/2-3/z +h.o.t. = -9z/2 +h.o.t.$

In particular, the principal part of the Laurent expansion is zero and hence there is a removable singularity at zero (residue $= 0$).

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Thanks Moritzplatz, makes a lot of sense, yes. –  Conan Wong Dec 16 '12 at 5:18
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Or simply Taylor ... $\sin(3z)=3z+o(z^2)$, so $$\lim_{z\to0}\frac{\sin(3z)-3z}{z^2}=\lim_{z\to0}\frac{o(z^2)}{z^2}=0\;.$$ Hence, the function extends holomorphically to $z=0$.

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Thanks wisefool - I guess this is similar to the Laurent series method. –  Conan Wong Dec 16 '12 at 5:18
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