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Let ${\pi }_{\Lambda }:\mathbb{C}\to \mathbb{C}/\Lambda$ and ${\pi }_{{\Lambda }^{\prime }}:\mathbb{C}\to \mathbb{C}/{\Lambda }^{\prime }$ be the universal covering maps of two one-dimensional complex tori. Every holomorphic map $f:\mathbb{C}/\Lambda \to \mathbb{C}/{\Lambda }^{\prime }$ has a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ satisfying ${\pi }_{{\Lambda }^{\prime }}\circ F=f\circ {\pi }_{\Lambda }$, and every such lift is an affine functionwith $a\Lambda \subseteq {\Lambda }^{\prime }$. Indeed, for $\lambda \in \Lambda$, the continuous function $F(z+\lambda )-F(z)$ takes values in the discrete set ${\Lambda }^{\prime }$ and is therefore constant. Hence $F^{\prime }$ is $\Lambda$-periodic. It is bounded on a compact fundamental parallelogram and therefore on the complex plane, so the Liouville theorem makes $F^{\prime }$ constant. If $f$ preserves the identity and the lift is chosen with $F(0)=0$, then $b=0$ and the lift is a linear map.

The complex modulus of $z=x+iy$ is $|z|=\sqrt{x^2+y^2}=\sqrt{z\overline{z}}$. It is the Euclidean norm of $(x,y)$ in the complex plane, and satisfies $|zw|=|z||w|$ and the triangle inequality.

An entire function is a holomorphic function whose domain is the whole complex plane.

A fundamental parallelogram for $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ is a translate ofIts translates tile the complex plane.

A bounded harmonic function on the complex plane is constant. More generally, an entire harmonic function bounded on either side is constant: write it as $\mathrm{Re}f$ by harmonic function as the real part of a holomorphic function, then apply the Liouville theorem to $e^f$ or $e^{-f}$.

The imaginary axis is the set of complex numbers $iy$ with $y\in \mathbb{R}$. It is the vertical coordinate axis in the complex plane.

If an elliptic function had no poles, it would be entire. It is bounded on the closure of a fundamental parallelogram, and periodicity then makes it bounded on the whole complex plane. The Liouville theorem would make it constant. Thus every nonconstant elliptic function has a pole.

Write $z=1+iy$ with $y\in \mathbb{R}$. In the complex plane, this is the vertical line through $1$. The complex exponential gives $e^z=e(\cos y+i\sin y)$, so its complex modulus is always $e$. Conversely, every point of that modulus has an complex argument $y$ and is obtained in this way. The image is the entire circle centered at zero with radius $e$, traversed repeatedly as $y$ varies; it is not a disk.

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its centre is $2\sqrt{\alpha \beta }$ on the real axis and its radius is $\alpha +\beta$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

A complex power series with radius of convergence $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the complex plane.

If $p$ is the zero polynomial, every coefficient vanishes, so $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,The ratio test therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the polynomial-coefficient power series has $R=1$ for every nonzero polynomial. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.