This was a question in an assignment that demands the use of Cauchy's Integral Formula in the questions. The question goes like this:
Integrate $\displaystyle{g(z)=\frac{e^z}{ze^z-2iz}}$ over $\displaystyle{C:\ \vert z\vert=0.5}$.
What I tried:
First, I noted that $\displaystyle{g(z)=\frac{e^z}{z\left(e^z-2i\right)}}$ so I could use the Cauchy's Integral Formula with $z_o=0$ and $\displaystyle{f(z)=\frac{e^z}{e^z-2i}}$. To do so, I have to make sure that $f(z)$ is analytic. So, I proceed like this:
As $z=x+iy$, so $$\begin{align} f(z)&=\frac{e^{x+iy}}{e^{x+iy}-2i}\\ &=\frac{e^x \cos y+ie^x\sin y}{e^x \cos y+ie^x\sin y-2i}\\ &=1+\frac{2i}{e^x \cos y+ie^x\sin y-2i}\\ &=1+\frac{2i\left(e^x \cos y-ie^x\sin y+2i\right)}{\left(e^x \cos y+ie^x\sin y-2i\right)\left(e^x \cos y-ie^x\sin y+2i\right)}\\ &=1+\frac{2ie^x\cos y+2e^x\sin y-4}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\\ &=\left[1+\frac{2e^x\sin y-4}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\right]+i\left[\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\right]\\ &=u+iv \end{align}$$ Hence $$u=1+\frac{2e^x\sin y-4}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\ \ \ \ \ \text{and}\ \ \ \ \ v=\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}$$ Now according to WolframAlpha, $$ \begin{align} \frac{\partial u}{\partial x}&=\frac{2e^x\sin y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{\left(2e^x\sin y-4\right)\left(2e^x\sin y\left(e^x\sin y-2\right)+2e^{2x}\cos^2y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}\\ \frac{\partial u}{\partial y}&=\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{\left(2e^x\sin y-4\right)\left(2e^x\cos y\left(e^x\sin y-2\right)-2e^{2x}\sin y\cos y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}\\ \frac{\partial v}{\partial x}&=\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{2e^x\cos y\left(2e^x\sin y\left(e^x\sin y-2\right)+2e^{2x}\cos^2 y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}\\ \frac{\partial v}{\partial y}&=-\frac{2e^x\sin y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{2e^x\cos y\left(2e^x\cos y\left(e^x\sin y-2\right)-2e^{2x}\sin y\cos y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2} \end{align} $$
Links for WolframAlpha: du/dx du/dy dv/dx dv/dy
Now clearly, these values for $\displaystyle\frac{\partial u}{\partial x}$, $\displaystyle\frac{\partial u}{\partial y}$, $\displaystyle\frac{\partial v}{\partial x}$ and $\displaystyle\frac{\partial v}{\partial y}$ do not satisfy the Cauchy-Riemann equations.
Did I do anything wrong during the solution? If not, what method should be used to solve the required integral?
Thanks for the attention.
If you consider $f$ defined in $U=D(0,r)$ where $\frac{1}{2}<r<\text{ln}(2)$ then $f(z)=\frac{e^z}{e^z-2i}$ is well-defined. This is true because $e^z-2i=0 \Rightarrow |e^z|=|2i| \iff e^{Re(z)}=2$ which is impossible in $U$, therefore $f$ is analytic once is the quotient of two analytic functions. Notice that $\frac{1}{2}<r$ implies $C\subset U$.
Finally by Cauchy formula, $\int_{C} \frac{e^z}{z(e^z-2i)} dz=\int_{C} \frac{f(z)}{z} dz=2\pi if(0)\text{ind}(0,C)=\frac{2\pi i}{1-2i}$ considering that $C$ goes counterclockwise.