Help with Complex Number (Locus/Argand Diagram) Hi please help with the following question:
"The complex number $\ z $  is given by $\ z=t+1/t $ where $\ t=r(cos\theta + isin\theta) $ find the equation of the locus of the point $\ P$ which represents $\ z $ on an Argand diagram in each of the following cases:


*

*$\ r=2 $ and $\theta $ varies

*$\theta = \pi/4 $ and $\ r $ varies"


I checked the answer, it state that for case (1) the resulting locus will be an ellipse while case (2) will be an Hyperbola. But I have no clue how to show it.
Please help and thank you.
 A: This I believe is the Joukowsky transform. I will refer to this transformation by the symbol $\omega$ rather than your $z$. Remember you can write
\begin{eqnarray}
t &=& r(\cos \theta + i \sin \theta) \\
   &=& re^{i \theta}  
\end{eqnarray}
From which under the transformation becomes
\begin{eqnarray}
\omega &=& t + \frac{1}{t} \\
       &=& re^{i \theta} +\frac{1}{re^{i \theta}} \\
       &=& re^{i \theta} + \frac{1}{r}e^{-i \theta} \\
       &=& r(\cos \theta + i \sin \theta) + \frac{1}{r}(\cos \theta - i \sin \theta) \\
       &=& (r+\frac{1}{r})\cos \theta + i(r-\frac{1}{r})\sin \theta 
 \end{eqnarray}
I'll leave you to fill in the blanks with the actual values in parts 1 and 2, but you should recognise the last line as a step to showing that under this transformation, the locus will map to an ellipse.
EDIT:
In order to see that this maps to an ellipse, let
\begin{eqnarray}
u &=& (r+\frac{1}{r})\cos \theta \\
v &=& (r-\frac{1}{r})\sin \theta \\
\end{eqnarray}
re-writing
\begin{eqnarray}
\frac{u}{(r+\frac{1}{r})} &=& \cos \theta \\
\frac{v}{(r-\frac{1}{r})} &=& \sin \theta 
\end{eqnarray}
Exploiting
\begin{equation}
\sin^{2} \theta +\cos^{2} \theta =1
\end{equation}
Gives an ellipse
\begin{equation}
\left(\frac{u}{(r+\frac{1}{r})}\right)^{2} + \left(\frac{v}{(r-\frac{1}{r})}\right)^{2} = 1
\end{equation}
