Proving $z^{2}+2z-1$ on the unit circle is a Cardioid. I want to prove 
$$f(z)=z^{2}+2z-1$$
restricted to $|z|=1$ is a Cardioid in the complex plane. 
I tried $z=e^{i\theta}$ which gave me 
$$f(z)=2e^{i\theta}(i\sin\theta+1)\equiv Re^{i\phi}$$
which looks very close if I let $\phi=\theta$, except for the fact that there is an $i$ in front of the $\sin \theta$.
How can I remove the factor of $i$ (or by other methods) to prove that $f(e^{i\theta})$ is a Cardioid in the complex plane? 
Follow up question: What type of polar curve is $z^{2}+2tz-t$ for $t\in (0,1)$ and how would I show this algebraically?
 A: You do not get the standard form because you are looking for a cardioid whose cusp is at the origin, and in this case it isn't.  Then where is it?
Take the imaginary part of $f(z)=z^2+2z-1$ and set it to zero, thus to find the real axis intercepts of the curve. Setting $x$ and $y$ as the real and imaginary parts of $f(z)$ this leads to
$2xy+2y=2(x+1)y=0$
The factor $x+1$ corresponds to a root $z=-1$ on the unit circle while the $y$ factor gives both $z=1$ and $z=-1$.  Thereby $z=-1$ is a multiple root and that will show up geometrically as the cusp if you have a cardioid.  Tge corresponding value of $f(z)$ at this cusp is then $-2$.
So to get the proper equation for a cardioid you need to look at polar coordinates based on the magnitude and argument of $f(z)+2$ rather than $f(z)$.  You should able to show that
$f(z)+2=(e^{i\theta}+1)^2$
and upon extracting the magnitude and modulus the proper equation for a cardioid comes out.
For the follow-up question the same method works.  In this case you find that $f$ becomes real when $x=-t$ corresponding to two points on the unit circle.  These two points correspond to a double root for $f$ where the curve crosses itself, and you use this point as the pole for your polar coordinate description.
Your polar equation with the proper location of the pole should then correspond to a limacon.  Geometrically, we may imagine that as $t$ drops below $1$ the cusp of the cardioid described earlier opens up to form the inner loop of the limacon, which grows and ultimately merges with the outer loop when $t$ reaches $0$ (thus $f(z)=z^2$).  The merged curve would then be two laps of the unit circle, which is what you would expect for $t=0$.
