what is the value of the contour integral oriented counterclockwise
$\displaystyle\oint_{|z|=1}\frac{e^z}{z^{10}}dz$ ??
my attempts : Applying Cauchy's Integral formula, you have that $\displaystyle\oint_{C_i}\dfrac{f(z)}{(z-a)^{n+1}}\,dz = \dfrac{2\pi i f(a)}{n!}$ if $a$ lies in the interior of the curve $C_i$, and $0$ otherwise.
here i got
$\displaystyle\oint_{|z|=1}\frac{e^z}{z^{10}}dz = \dfrac{2\pi i f(0)}{9!}=\dfrac{2\pi i }{9!} where, e^0 =1$
is my answer is correct or not pliz verified its
Cauchy's integral formula actually states $$\int_C \frac{f(z)}{(z-a)^{n+1}}\,dz=2\pi i\frac{f^{(n)}(a)}{n!}$$ when the contour $C$ winds round the point $a$ once in the anticlockwise direction. Fortunately you are dealing with $f(z)=e^z$ which is its own derivative, and so fortuitously you have the correct answer.
Another approach is to write $$\frac{f(z)}{z^{10}}=\frac1{z^{10}}+\frac1{z^9}+\cdots +\frac1{9!z}+\frac1{10!}+\cdots$$ and observe that $$\frac{f(z)}{z^{10}}-\frac1{9!z}$$ has an antiderivative, and so this integrates to zero on the closed contour $C$.