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I have working through past exam questions and I think I have the hang of the Cauchy integral formula and the extended formula... but am a little stuck with how to work these examples out... and the denominators aren't of the form $(z-z_0)^n$

Any help in how to solve these types of questions would be greatly appreciated...

$(i) \displaystyle\int_\gamma \dfrac{\sin z}{z^4-16}$ (with $\gamma$ in the unit circle in $\mathbb{C}$ traversed in the anti-clockwise direction)

$(ii)\displaystyle\int_\gamma \dfrac{\cos z}{z(z^2-8)}$ (with $\gamma$ being the circular contour given by $\gamma(t)=e^{it}$ for $1 \leq t \leq 2\pi$

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    $\begingroup$ I take it you mean $0 \le t \le 2 \pi$. $\endgroup$
    – Ron Gordon
    Commented May 3, 2013 at 13:57
  • $\begingroup$ @RonGordon No, in the past paper it is says $1\leq t \leq 2\pi $ $\endgroup$ Commented May 3, 2013 at 14:22
  • $\begingroup$ That makes absolutely no sense. I mean, it is a problem, but what of it? $\endgroup$
    – Ron Gordon
    Commented May 3, 2013 at 14:28

2 Answers 2

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$(i)~z^4-16=0\iff z=\pm2,\pm2i.$

All of $\pm2,\pm2i$ lie outside $\gamma.$ So $\dfrac{\sin z}{z^4-16}$ is analytic everywhere on and inside $\gamma.$

Consequently, by Cauchy-Goursat theorem $\displaystyle\int_\gamma \dfrac{\sin z}{z^4-16}=0.$ (Similar Question)

$(ii) \dfrac{\cos z}{z(z^2-8)}$ is analytic everywhere on and inside $\gamma$ except at $0.$ Note $\dfrac{\cos z}{z(z^2-8)}$

$=-\dfrac{1}{8z}\left(1+\dfrac{z^2}{2!}+\dfrac{z^4}{4!}+...\right)\left(1-\dfrac{z^2}{8}\right)^{-1}$$=-\dfrac{1}{8z}\left(1+\dfrac{z^2}{2!}+\dfrac{z^4}{4!}+...\right)\left(1+\dfrac{z^2}{8}+...\right)$$=...$ is the Laurent series expansion of $\dfrac{\cos z}{z(z^2-8)}$ around $0.$

So $\text{Res}_{z=0}\dfrac{\cos z}{z(z^2-8)}=-\dfrac{1}{8}.$

Consequently, due to the Residue theorem, $\displaystyle\int_\gamma \dfrac{\cos z}{z(z^2-8)}=2\pi i\times\left(-\dfrac{1}{8}\right)=-\dfrac{\pi i}{4}.$

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  • $\begingroup$ Thank you. I've worked it all out now, thanks for your help... Just one last thing I have come across another question $\displaystyle\int_\gamma \dfrac{e^ {iz}}{(z+2i)(z-2i)}dz$ (LEt $\gamma$ be the contour given by $\gamma(t)=e^{it}$ for $1\leq t\leq 0$) Surely this =0, as 2i and -2i lie outside the circle? Im just a little confused by this as its worth less marks on the paper than other contour integrals which need a lot more work? $\endgroup$ Commented May 3, 2013 at 14:50
  • $\begingroup$ $\gamma=\{e^{it}:0\le t\le 1\}$ is not a circle. Rather an arc of the unit circle. You should define $t$ over $[0,2\pi].$ $\endgroup$ Commented May 3, 2013 at 15:04
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(i): what are the roots of the denominator? Where do they lie with respect to the interior of $\gamma$?

Note that the roots of $z^4-16$ all have magnitude of $2$.

(ii): same questions. Note however that $z=0$ is a root and therefore inside $\gamma$. IN that case, the residue of that pole is

$$i 2 \pi \frac{1}{-8} = -i \frac{\pi}{4}$$

The other roots have magnitude $2 \sqrt{2}$; what does that mean?

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    $\begingroup$ so for part (i) $\displaystyle\int_\gamma \dfrac{\sin z}{(z-2)(z+2)(z-2i)(z+2i)}$ =0 as 2,-2,2i and -2i lie outside the circle? $\endgroup$ Commented May 3, 2013 at 14:12
  • $\begingroup$ @Mathsstudent147: yes $\endgroup$
    – Ron Gordon
    Commented May 3, 2013 at 14:28
  • $\begingroup$ Ron, what would be different in part (i) if we changed the contour to, say $|z| = 3$? $\endgroup$ Commented Oct 23, 2013 at 18:40
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    $\begingroup$ @TheChaz2.0: now the poles are interior to the contour so you'd have to compute the residues at all the poles and sum. $\endgroup$
    – Ron Gordon
    Commented Oct 23, 2013 at 21:17

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