I have an integral $$\oint_{|z|=1}\frac{e^z}{z}\,\mathrm dz.$$ I defined $g: [0,2\pi]$, $g(t)=e^{it}$. The integral then becomes. $$\int_0^{2\pi}e^{e^{it}}\mathrm dt.$$ I used the property that states that the value of a function at the center of a closed circle is the same as the mean value of that same function at the circumference (comes from Cauchy's integral formula).
My answer was $i2\pi$. The book comes with no answers. Is it correct? Thank you.
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You're right, but why not Cauchy's Integral Formula?
$$\oint_{|z|=1}\frac{e^z}{z}\,\mathrm dz$$
$$ = \oint_{|z|=1}\frac{e^z}{(z-w)^{n+1}}\,\mathrm dz$$
$$ = 2 \pi i \frac{1}{n!}f^{n}(w) = 2 \pi i$$
where $n=0,w=0 \in int(\{|z|=1\})$ and $f(z)=e^z$, which is holomorphic in $int(\{|z|=1\})$ because $f: \mathbb C \to \mathbb C$ is entire.
Yes, it is correct, but you don't need to convert your integral to $\int_0^{2\pi}ie^{e^{it}}\,\mathrm dt$ (note that you missed the first $i$). Just apply directly Cauchy's integral formula:$$\frac1{2\pi i}\oint_{\lvert z\rvert=1}\frac{e^z}z\,\mathrm dz=e^0=1.$$