Show that the tranformation $ w = z + \frac{1}{z}$ transform r = constant in the z plane into a family of ellipses in the w plane.

Question

Here we have,

$$w = z + \frac{1}{z}$$

$$\therefore u + iv = x + iy + \frac{1}{x + iy}$$

On solving and comparing the real and imaginary parts we get,

$$ u = x(1 + \frac{1}{x^2 + y^2})$$ and

$$ v = y(1 - \frac{1}{x^2 + y^2})$$

What's next from here?

Answer 1

Let $ z=x+iy$ and $w=u+iv$. Then, $w=z+\frac1z$ becomes

$$u+iv= x+iy +\frac 1{x+iy}= x+iy +\frac{x-iy}{r^2}$$

Equate the real and imaginary parts, respectively, to get

$$x=\frac{r^2 u}{r^2+1},\>\>\>\>\>y=\frac{r^2 v}{r^2-1}$$

and substitute above into $x^2+y^2=r^2$ to obtain

$$\frac{r^2u^2}{{(r^2+1)^2}}+ \frac{r^2v^2}{{(r^2-1)^2}}=1$$

which represents an ellipse centered at origin, with the major and the minor axes of $\frac{r^2+1}{r}$ and $\frac{r^2-1}{r}$, respectively.

Answer 2

Slightly differently:

Given that $|z|=r, (r \ne 1)$, the $$w=u+iv=z+1/z=z+\bar z/(z\bar z)=z +\bar z/r^2=x(1+1/r^2)+iy(1-1/r^2)$$ $$\implies x=u/(1+1/r^2), y=v/(1-1/r^2)$$ As $x^2+y^2=r^2$ we get ellipse in $(u,v)$ space as $$\frac{u^2}{(r^2+1)^2}+\frac{v^2}{(r^2-1)^2}=1, r\ne 1.$$