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Given $D=\{r(\sin(\alpha)+i\cos(\alpha))\mid r>0\}= \{e^{x+ i y} \mid x\in \mathbb{R} \}$ (for fixed a $y$ and $\alpha$) i need to find $f(D)$ where $f$ is the the joukowski transform: $$f(z)=\frac{1}{2}(z+\frac{1}{z})$$

After playing around with it i got to ($y=\theta$) $$\frac{\sin\left(\theta\right)}{2}\left(e^{x}-e^{-x}\right)+i\frac{\cos\left(\theta\right)}{2}\left(e^{x}+e^{-x}\right)$$

I am afraid i got somewhat lost, Any suggestions would be appreciated.

1 Answers 1

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turns out that all i needed was the following:

$$\frac{\sin\left(\theta\right)}{2}\left(e^{x}-e^{-x}\right)+i\frac{\cos\left(\theta\right)}{2}\left(e^{x}+e^{-x}\right)$$

if we take $u=\frac{\sin\left(\theta\right)}{2}\left(e^{x}-e^{-x}\right)$ and $v=\frac{\cos\left(\theta\right)}{2}\left(e^{x}+e^{-x}\right)$ then i get that: $$ \frac{v^2}{sin^2(y)} - \frac{u^2}{cos^2(y)}=1$$ Which means i get a hyperbola!