I have a question Let γ be the circular contour, positively oriented, with centre 0 and radius 8. Compute the following integral

$$\int \frac{5!\cos(z)}{(z-2\pi )^6}$$

I used the formula $$f^n(a)= \frac{n!}{2\pi i}\int \frac{f(z)}{(z-a)^{n+1}}$$

So I wrote $$f^n(a)=\frac{5!}{2\pi i} \int \frac{\cos z}{(z-2\pi)^6}$$

= $$\frac{5!}{2 \pi i} . \frac{2 \pi i}{5!}.f^5(2 \pi)=1$$

I'm new to Cauchy integrals so I'm not sure if I did this right help would be really appreciated.

  • $\begingroup$ Is $f^{(5)}(2\pi)=1$? $\endgroup$ – 민찬홍 Nov 18 '17 at 14:43
  • $\begingroup$ math.stackexchange.com/questions/2526074/… $\endgroup$ – Guy Fsone Nov 18 '17 at 14:52
  • $\begingroup$ I wrote f(z) as cosz and I f^5 (cos(z))=cosz so it would be cos 2pi if im not mistaken? $\endgroup$ – Sarah Angel Nov 18 '17 at 14:53
  • $\begingroup$ @SarahAngel No it is $-\sin z $. $\endgroup$ – 민찬홍 Nov 18 '17 at 15:02

There's a problem concerning the notation: it's $f^{(n)}$, not $f^n$. Besides, when you write:$$f^{(n)}(a)=\frac{5!}{2\pi i}\int\frac{\cos z}{(z-2\pi)^6},$$

  • $\mathrm dz$ is missing;
  • $\gamma$ is missing;
  • you don't tell us what $a$ is.

But the idea is correct.

  • $\begingroup$ and the parenthesis in the exponent means that many number of times differentiated, or am i tired? $\endgroup$ – mathreadler Nov 18 '17 at 15:40
  • $\begingroup$ @mathreadler Yes, that's what it means. $\endgroup$ – José Carlos Santos Nov 18 '17 at 15:41
  • $\begingroup$ I just wanted to clarify it as no one else had pointed it out. $\endgroup$ – mathreadler Nov 18 '17 at 15:42
  • $\begingroup$ That was all the information given to me in the question $\endgroup$ – Sarah Angel Nov 18 '17 at 16:03

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