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http://gyazo.com/8ef04b854bc3bbfb6b55a9af45e51fdc.png

Since $f(z)$ not differentiable at $0$, isolated singularity at $z = 0$. By expanding the Laurent series and looking at the first term, I got a pole of order $3$ (is there a more proper way to get this?).

For $a_{-1}$ I got $-8\pi{i}/3$. Now I don't know how to do the last question. Do I use a circle of radius $3$ for C? Wouldn't that just give me $a_{-1}$ back? Help would be much appreciated, thanks in advance.

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You have an entire function $f$ with a well-known Taylor series. To obtain the Laurent series of $\dfrac{f(z)}{z^k}$ about $0$, you just take the Taylor series of $f$ and divide that by $z^k$, there is no better way. Here, we have

$$\frac{\sin (2z)}{z^4} = \frac{1}{z^4}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(2z)^{2n+1} = \sum_{n=0}^\infty \frac{(-1)^n 2^{2n+1}}{(2n+1)!} z^{2n-3}.$$

For a_-1 I got -8pi*i/3

There is no $\pi$ in sight in the Laurent series, and the coefficients are real. The coefficient $a_{-1}$ is obtained by picking the term for $n = 1$ in the series above, that yields

$$a_{-1} = \frac{(-1)^1 2^3}{3!} = -\frac{4}{3}.$$

What you have written down is the answer to the last question. Since $\dfrac{\sin (2z)}{z^4}$ has no other singularities,

$$\oint_C \frac{\sin (2z)}{z^4}\,dz$$

is, for $C$ a contour winding once around the origin, $2\pi i$ times the residue in $0$, that is, $2\pi i a_{-1}$, irrespective of the chosen contour, as long as the origin doesn't lie on the contour, and it winds once around the origin. Generally,

$$\oint_C \frac{\sin (2z)}{z^4}\,dz = 2\pi i \; n(C,0) a_{-1},$$

where $n(C,0)$ is the winding number of $C$ around the origin.

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  • $\begingroup$ Appreciate it, finding a_-1 was simpler than I thought. $\endgroup$ – Freud Feb 10 '14 at 18:56
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Write $$f(z)=({sin(2z)}/z)z^{-3}.$$ Then the first factor is analytic in $z=0$ and tends to 2 as $z\rightarrow{0}$. The integral $$∮_{C}dzsin(2z)/z⁴=∮_{C}dz{\{}sin(2z)/z{\}}1/(z³)$$ vanishes. This is usually proven by using polar coordinates and integrating along a circle around the pole. Put $z=r{\text{exp}}[iϕ]$. Then $$∮_{C}dz((sin(2z))/z)(1/(z³))∼2∮_{C}dz(1/(z³))=2{\int}{_0}^{2π}dϕir{\text{exp}}[iϕ](1/(r³{\text{exp}}[3iϕ]))=((2i)/(r²)){\int}{_0}^{2π}dϕ{\text{exp}}[-2iϕ]=0$$ Edit: As pointed out by @Daniel Fisher and @DonAntonio I made a mistake. Here is an improved (and hopefully correct version.$$ f(z) = \frac{sin(2z)}{z^4}=\frac{1}{z^4}{\{}2z-\frac{(2z)^3}{3!}+\frac{(2z)^5}{5!}-…{\}}=\frac{2}{z^3}-\frac{4}{3z}+\frac{4}{15}z-…, = \frac{2}{z^3}-\frac{4}{3z}+{\{}f(z)-\frac{2}{z^3}+\frac{4}{3z}{\}}=\frac{2}{z^3}-\frac{4}{3z}+g(z)$$where $g(z)$ is analytic in $0$. Now $$∮_{C}dzf(z)=∮_{C}dz\frac{sin(2z)}{z^4}=∮_{C}dz{\{}\frac{2}{z^3}-\frac{4}{3z}+g(z){\}}=∮_{C}dz{\{}\frac{2}{z^3}-\frac{4}{3z}{\}}=-\frac{4}{3}∮_{C}dz\frac{1}{z}=-\frac{4}{3}2πi=-\frac{8πi}{3}$$

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  • $\begingroup$ The integral does not vanish, it's the quadratic term of $\dfrac{\sin (2z)}{z}$ that in conjunction with the $1/z^3$ produces a nonzero result. $\endgroup$ – Daniel Fischer Feb 10 '14 at 11:23
  • $\begingroup$ That $\;\sim\;$ sign in the line before the last one is one of the oddest things I've ever seen in complex integration...and it leads to a wrong result. $\endgroup$ – DonAntonio Feb 10 '14 at 11:45
  • $\begingroup$ @DanielFischer, $DonAntonio. You are right, I have been too hasty. $\endgroup$ – Urgje Feb 10 '14 at 14:44

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