Conformal mapping of cardoid $r = \rho ( 1 + \cos \theta )$

Question

Where does the cardoid $r = \rho ( 1 + \cos \theta )$ map in the $w$ plane, by function $w = \sqrt z$ ?

Answer 1

Note that point $\left(r,\theta\right)$ in the $w$ plane corresponds to the point $\left(r^2,2\theta\right)$ in the z plane. So $r=p\left(1+cos\theta\right)$ becomes $r^2=p\left(1+\cos 2\theta\right).$ (It's not the other way around, think about it for a moment.)

Replacing $r^2$ with $x^2+y^2$, $\cos 2\theta$ with $\displaystyle\frac{x^2-y^2}{x^2+y^2}$, and then simplifying, we get that $\displaystyle\left(x\pm\sqrt{\frac{p}{2}}\right)^2+y^2=\frac{p}{2}$

This is the same as two circles radius $\displaystyle\sqrt{\frac{p}{2}}$ centered at $\left(\pm\displaystyle\sqrt{\frac{p}{2}},0\right)$.

Answer 2

A parametric equation for the cardioid is : $$x=\rho \cos (t)(1+\cos (t));y=\rho \sin (t)(1+\cos (t))$$ Or : $z=\rho e^{it}(1+\cos(t))$ You apply the function $w$ : $$w(z)=\sqrt{\rho(1+\cos t)}e^{\frac{it}{2}}=\sqrt{\rho(1+\cos t)}(\cos (\frac{t}{2})+ i\sin (\frac{t}{2}))$$ So the parametric equation of the image of the caridoid is : $$x=\sqrt{\rho(1+\cos t)}\cos (\frac{t}{2});y=\sqrt{\rho(1+\cos t)}\sin (\frac{t}{2})$$ I will let you study and draw this last curve but it has the shape of an "$8$".