calculate $\oint_{|z|=1}z^{2018}e^{\frac{1}{z}}\sin\frac{1}{z}\text{dz}$ my try:
$ \begin{array}{c} \oint_{|z|=1}z^{2018}e^{\frac{1}{z}}\sin\frac{1}{z}\text{dz}\\ \oint_{|z|=1}z^{2018}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(\frac{1}{z}\right)^{n}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(\frac{1}{z}i\right)^{n}-\left(-\frac{1}{z}i\right)^{n}}{2in!}\text{dz}\\ \oint_{|z|=1}z^{2018}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{z^{-n}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(-zi\right)^{-n}-\left(zi\right)^{-n}}{2in!}\text{dz}\\ \oint_{|z|=1}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{z^{-n+2018}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(-zi\right)^{-n}-\left(zi\right)^{-n}}{2in!}\text{dz}\\ \oint_{|z|=1}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{z^{-n+2018}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(-zi\right)^{-n}-\left(zi\right)^{-n}}{2in!}\text{dz}\\ {\displaystyle \mathop{Res\sum_{0}^{\infty}}}\frac{z^{-n+2018}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(-zi\right)^{-n}-\left(zi\right)^{-n}}{2in!}\text{}\\ {\displaystyle \mathop{Res\sum_{0}^{\infty}}}\frac{z^{-n+2018}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(z\right)^{-n}(-i)^{-n}-\left(z\right)^{-n}(i)^{n}}{2in!}\text{}\\ {\displaystyle \mathop{Res\sum_{0}^{\infty}}}\frac{z^{-2n+2018}\left((-i)^{-n}-i^{n}\right)}{2i(n!)^{2}}\text{}\\ -2n+2018=-1\\ n\notin\mathbb{N}\\ {\displaystyle \mathop{Res\sum_{0}^{\infty}}}\frac{z^{-2n+2018}\left((-i)^{-n}-i^{n}\right)}{2i(n!)^{2}}=0 \end{array}$
$\oint_{|z|=1}z^{2018}e^{\frac{1}{z}}\sin\frac{1}{z}\text{dz}=0$ edit: I took the advise in the comment now I get: $ {\displaystyle \mathop{Res\sum_{0}^{\infty}}}\frac{z^{-n+2018}}{n!}{\displaystyle \mathop{\sum_{0}^{\infty}}}\frac{\left(z\right)^{-k}(-i)^{-k}-\left(z\right)^{-k}(i)^{k}}{2ik!}\text{}\\{\displaystyle \mathop{Res\sum_{0}^{\infty}}}\frac{z^{-n-k+2018}\left((-i)^{-k}-i^{k}\right)}{2i(n!)(k!)}\text{}\\-n-k+2018=-1\\n+k=2019\\{\displaystyle \mathop{Res\sum_{n+k=2019;n,k=0}}}\frac{z^{-n-k+2018}\left((-i)^{-k}-i^{k}\right)}{2i(n!)(k!)}$ which I don't know to sum
${\displaystyle \mathop{Res\sum_{k=0}^{2019}}}\frac{z^{-1}\left((-i)^{-k}-i^{k}\right)}{2i((2019-k)!)(k!)}={\displaystyle \mathop{Res\sum_{k=0}^{2019}}}\frac{\left((-i)^{-k}-i^{k}\right)}{2i((2019-k)!)(k!)}$
If inside the unit disc is too nasty, we can try to move it to outside the unit disc. In other words, integrate it on the other side of the Riemann sphere using the chart $w=\frac1z$.
So, with $w=\frac1z$, the unit circle $\lvert z\rvert=1$ is $\lvert w\rvert=1$, in the opposite direction. Everything else is just what you expect: $$ \int_{S^1}z^{2018}\exp(z^{-1})\sin(z^{-1})\,\mathrm{d}z =-\int_{S^1}w^{-2018}\exp(w)\sin(w)\cdot -w^{-2}\,\mathrm{d}w. $$ Now you can calculate the RHS much more readily: $\exp(w)\sin(w)$ is holomophic and we have $$ \exp(w)\sin(w)=\frac1{2i}\left[\exp((1+i)w)-\exp((1-i)w)\right] $$ so \begin{align*} &=\frac{1}{2i}\int_{S^1}\frac{\exp((1+i)w)-\exp((1-i)w)}{w^{2020}}\,\mathrm{d}w \\ &=\pi\cdot\text{coefficient of }w^{2019}\text{ in }\big[\exp((1+i)w)-\exp((1-i)w)\big]\\ &=\pi\frac{(1+i)^{2019}-(1-i)^{2019}}{2019!} \end{align*} and it is not hard to calculate powers of $1\pm i$ using $(1\pm i)/\sqrt{2}$ is an 8-th root of unity.