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I was going over some practice problems

T/F: If $C$ is the circle in $\mathbb{C}$ of radius $10$ centered at $z=2i$ with positive orientation and $$f(z) = \frac{\cos(4z)}{z}$$ then $$ \oint_{C} f(z) \mathrm{d}z = 2\pi i\, \mathrm{}.$$

I used the Cauchy Integral Theorem, with $z_0= 0$ and $f(z)= \cos(4z)$. That gives the value as $2\pi i$, but the answer is given as false? What mistake am I making?

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  • $\begingroup$ @Fakemistake - OP got that answer, but was told the answer was something else. $\endgroup$ Apr 16, 2020 at 7:30

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Okay, so let's try another technique to verify your answer. Your curve $C$ can be parameterized by $z(t) = 10e^{it} + 2i$ for $t \in [0,2\pi)$. Then your integral becomes

$$\int_0^{2\pi} \frac{\cos(40e^{it} + 8i)}{10e^{it} + 2i} \cdot 10ie^{it} \cdot dt$$

Obviously this isn't something you want to bother calculating through this method. But WolframAlpha can approximate it (as seen here), and returns an answer of about $6.28319i$. This obviously is quite close to $2\pi i$.

With this in mind, and the fact that I see nothing wrong with your approach (though you shouldn't say $f(z) = \cos(4z)/z$ and then also have $f(z) = \cos(4z)$ represent the numerator), I assume the answer is in error.

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