# Why isn't $\oint_{C} f(z) = 2\pi i\, \mathrm{d}z$?

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?

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

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.