Is the centrifugal force a real force?
Does centrifugal force exist?
Let $\mathbb{S} = \{ (X,Y,Z) \in \mathbb{R}^3 :X^2 + Y^2+Z^2=1\}$ denote the Riemann sphere. The following problem is taken from Gamelin "Complex Analysis" chapter one:     What is the image of the spherical cap $ C := \{(X,Y,Z) \in \mathbb{S} : A \leq X \leq 1\}$, where $A$ is some fixed real number between $-1$ and $1$, under the stereographic projection $ \pi(X,Y,Z) = \frac{X+iY}{1-Z}$?   Here is my work so far:  ( The case $A=1$ is trivial, so assume $A &lt;1$)  We want to find $\pi(C) = \{z \in \mathbb{C} \mid \exists(X,Y,Z) \in C: \pi(X,Y,Z) = z \}= \{z \in \mathbb{C} \mid \pi^{-1}(z) \in C\}$, where the last equality holds by $\pi$ being a bijection. Write $z=x+iy$.  Here, $ \pi^{-1}(z) \in C\iff A\leq \frac{2x}{1+x^2+y^2} \leq 1 \iff (\star) :A \leq \frac{2x}{1+x^2+y^2}$, since the inequality $2x \leq 1+x^2+y^2$ holds for all $x,y \in \mathbb{R}$.  $A=0$: $( \star)$ reduces to $ 0 \leq x$, so in this case the image is $\{z \in \mathbb{C} \mid \Re(z) \geq 0 \}$.   $A &gt;0$: Since $A \neq 0$ in $(\star)$ we have,  $ \frac{2x}{1+x^2+y^2} \geq A \iff Ax^2-2x+Ay^2+A \leq 0 \iff (x-\frac{1}{A})^2 +y^2 \leq \frac{1}{A^2}-1$. Thus the image is the interior and boundary of the circle centered at $\frac{1}{A}$ with radius $\sqrt{\frac{1}{A^2}-1}$. Note $\frac{1}{A^2}-1 \geq 0$ as $A \in (0,1)$.  $A &lt;0$ : By symmetry and the above, here the image is $ \mathbb{C} / \{ $ interior and boundary of circle centered at  $  \frac{1}{A}$ with radius $\sqrt {\frac{1}{A^2}-1} \}$.  Is my working correct? Many Thanks! 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