$\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\suchthat}{\;\big\vert\;}$
Let $ \mathbb{R}^{2} = \set{(x,y,z)\in\mathbb{R}^{3} \suchthat z=-1} $ be identified with the complex plane $ \mathbb{C} $ by setting $(x,y,-1)=x+iy=\zeta\in\mathbb{C}$. Let $ P: \mathbb{C}\to\mathbb{C} $ be the complex polynomial \begin{equation*} P(\zeta) = a_{0}\zeta^{n} + a_{1}\zeta^{n-1} + \cdots + a_{n}, \quad a_{0}\neq 0,\, a_{i}\in\mathbb{C},\, i=0,\dots,n. \end{equation*} Denote by $ \pi_{N} $ the stereographic projection of $ S^{2}=\set{(x,y,z)\in\mathbb{R}^{3} \suchthat x^{2}+y^{2}+z^{2}=1 } $ from the North pole $ N=(0,0,1) $ onto $ \mathbb{R}^{2} $. Prove that the map $ F: S^{2}\to S^{2} $ given by \begin{align*} F(p) &= \pi_{N}^{-1}\circ P \circ \pi_{N}(p), \quad \text{if } p\in S^{2}-\set{N}, \\ F(N) &= N \end{align*} is differentiable.
My attempt
By using Do Carmo's hint, $ F $ is differentiable in $ S^{2}-\set{N} $ as a composition of differentiable maps. To prove that $ F $ is differentiable at $ N $, consider the stereographic projection $ \pi_{S} $ from the South pole $ S=(0,0,-1) $ and set $ Q=\pi_{S} \circ F \circ \pi_{S}^{-1}: U \subset \mathbb{C}\to\mathbb{C} $ (of course, we are identifying the plane $ z=1 $ with $ \mathbb{C} $)
Now, I will show that $ \pi_{N}\circ\pi_{S}^{-1}: \mathbb{C}-\set{0}\to\mathbb{C} $ is given by $ \pi_{N}\circ\pi_{S}^{-1}(\zeta) = 1/\overline{\zeta} $.
Let $ \mathbf{u}=(u,v,1) $ be a point in plane $ z=1 $ (i.e., $ \mathbb{C} $); $ \ell_{S}(t) $ be the line through $ S $ with director vector $ \mathbf{u}-S=(u,v,2) $, then the parametrization of $ \ell_{S} $ is \begin{equation*} \ell_{S}(t) = (0,0,-1) + t(u,v,2) = (tu,tv,2t-1) \end{equation*} and the intersection point $ (x,y,z) $ between $ \ell_{S} $ and $ S^{2} $ (distinct of $ S $, i.e., $ t \neq 0 $) is given by the value of $t$ that satisfies \begin{equation*} (tu)^{2} + (tv)^{2} + (2t-1)^{2}= 1 \quad \implies \quad t^{2}u^{2} + t^{2}v^{2} + 4t^{2} - 4t + 1 = 1 \quad \implies \quad tu^{2} + tv^{2} + 4t = 4 \end{equation*} then \begin{equation*} t=\dfrac{4}{u^{2} + v^{2} + 4} \end{equation*} yields \begin{equation*} (x,y,z) = \left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{4-u^{2}-v^{2}}{u^{2} + v^{2} + 4} \right) \end{equation*} therefore \begin{equation} \pi_{S}^{-1}(u,v) = \left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{4-u^{2}-v^{2}}{u^{2} + v^{2} + 4} \right) \tag{1} \end{equation} Now, to calculate $ \pi_{N}\circ\pi_{S}^{-1} $, take (1) as a point in $ S^{2} $ and let $ \ell_{N}(t) $ be the line trough $ N $ with director vector $ \pi_{S}^{-1}-N $, then \begin{align*} \ell_{N}(t) &= (0,0,1) + t\left[ \pi_{S}^{-1} - (0,0,1) \right] \\[2ex] &= (0,0,1) + t\left[ \left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{4-u^{2}-v^{2}}{u^{2} + v^{2} + 4} \right) - (0,0,1) \right] \\[2ex] &= (0,0,1) + t\left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{-2(u^{2}+v^{2})}{u^{2} + v^{2} + 4} \right) \end{align*} And for this point to belong to the plane $ z=-1 $ (i.e., $ \mathbb{C} $) clearly we need that \begin{equation*} \dfrac{-2(u^{2}+v^{2})}{u^{2} + v^{2} + 4}t+1=-1 \quad \implies \quad t=\dfrac{u^{2} + v^{2} + 4}{u^{2} + v^{2}} \end{equation*} thus \begin{equation*} (x,y,-1) = \left( \dfrac{4u}{u^{2}+v^{2}}, \dfrac{4v}{u^{2}+v^{2}}, -1 \right) \end{equation*} therefore \begin{equation} \pi_{N}\circ\pi_{S}^{-1}(u,v) = \left( \dfrac{4u}{u^{2}+v^{2}}, \dfrac{4v}{u^{2}+v^{2}}, -1 \right) \tag{2} \end{equation} where $ (u,v-1) = u + iv = \zeta $. \ But the inverse of conjugate of $ \zeta $ should be \begin{equation} 1/\overline{\zeta} = \left( \dfrac{u}{u^{2}+v^{2}}, \dfrac{v}{u^{2}+v^{2}}, -1 \right) \tag{3} \end{equation}
My Question. Where is my mistake? Do Carmo's book says the equation should be like (3). I'm confused.
Next step is conclude that \begin{equation*} Q(\zeta) = \dfrac{\zeta^{n}}{\overline{a_{0}} + \overline{a_{1}}\zeta + \cdots + \overline{a_{n}}\zeta^{n}} \end{equation*} We have the next relations (if $ F \neq N $) \begin{align*} Q(\zeta) &= \left( \pi_{S}\circ F \circ\pi_{S}^{-1} \right) (\zeta) = \left( \pi_{S}\circ \pi_{N}^{-1} \circ P \circ\pi_{N} \circ\pi_{S}^{-1} \right) (\zeta) \\ &= \left( \pi_{S}\circ \pi_{N}^{-1} \circ P \right) \circ \left( \pi_{N}\circ\pi_{S}^{-1}(\zeta) \right) \\ &= \left( \pi_{S}\circ \pi_{N}^{-1} \circ P \right) (1/\overline{\zeta}) \\ &= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \circ P(1/\overline{\zeta}) \\ &= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \left( a_{0}\left(1/\overline{\zeta}^{n}\right) + a_{1}\left(1/\overline{\zeta}^{n-1}\right) + \cdots + a_{n} \right) \\ &= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \left( \left( 1/\overline{\zeta}^{n} \right)\left( a_{0}+a_{1}\overline{\zeta} + \cdots + a_{n}\overline{\zeta}^{n} \right) \right) \end{align*} Since $ \pi_{S}\circ \pi_{N}^{-1} = \left( \pi_{N}\circ\pi_{S}^{-1} \right)^{-1} $ should be $ \pi_{S}\circ \pi_{N}^{-1} = 1/\overline{\zeta} $ as well. Thus \begin{align*} Q(\zeta) &= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \left( \left( 1/\overline{\zeta}^{n} \right)\left( a_{0}+a_{1}\overline{\zeta} + \cdots + a_{n}\overline{\zeta}^{n} \right) \right) \\ &= \dfrac{\zeta^{n}}{\overline{a_{0}} + \overline{a_{1}}\zeta + \cdots + \overline{a_{n}}\zeta^{n}} \end{align*} Hence, $ Q $ is differentiable at $ \zeta=0 $. Thus, $ F = \pi_{S}^{-1}\circ Q \circ\pi_{S} $ is differentiable at $ N $.
DoCarmo's intent is that you project onto the plane $z=0$, not the plane $z=-1$. (This is the standard construction of stereographic projection.)