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Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that