I want to find the image of the complex transformation $f(z)=w=2z+\frac{1}{z}$ on the exterior of the unit circle in the complex plane. This is clearly a conformal mapping except at the points $z=0$ where the function has a pole and at the points $z=\pm \frac{1}{\sqrt{2}}$ where $f'(z)=0$. So we can make use of the fact that we know for conformal mappings: $\partial f(D)=f(\partial D)$.
Hence, we should just have to find the image of the boundary and a single point. To do this, I started by noting that: $$ \partial D = \{z=e^{i\theta}:0\leq\theta<2\pi\} $$ I then found the real and imaginary parts of the function $f(z)=w=u+iv$: $$ w = 2z+\frac{1}{z} = 2x+i2y+\frac{x-iy}{x^2+y^2} = (2x+\frac{x}{x^2+y^2}) + i(2y - \frac{y}{x^2+y^2})=u+iv $$ However, here is where I'm starting to confuse myself. On the boundary of $D$, we know that $x^2+y^2=r^2=1$ and we can convert our cartesian coordinate to polar, so we have: $$ w = u+iv = (3x)+i(y) = (3\cos(\theta))+i(\sin(\theta)) $$ Then it follows that $-3\leq u\leq3$ and $-1\leq v\leq1$. But aren't Joukowski transformations supposed to map the exterior of circles to the exterior of an ellipse? Am I missing something or oversimplifying somewhere?