## Geometry ### The Extended Complex Plane * The **Riemann sphere** is the sphere in $\mathbb{R}^3$ with radius $\frac{1}{2}$, centered at $\left( 0,0, \frac{1}{2} \right)$: $\Sigma=\left\{ (x,y,z) \in \mathbb{R}: x^2+y^2+\left( z-\frac{1}{2} \right)^2=\frac{1}{4} \right\}$ * The **extended complex plane** is the set $\mathbb{\tilde{C}} =\mathbb{C} \cup \{ \infty \}$. * The **stereographic projection** maps the extended complex plane to the Riemann sphere. * Geometrically, it maps a point $z=x+yi \in \mathbb{C}$ to the intersection point of the Riemann sphere and the line joining $(x,y,0)$ and the "north pole" $(0,0,1)$. The point $\infty$ is mapped to the north pole. ### Circlines * Any circle on $\Sigma$ that does not pass the north pole corresponds to a circle on $\mathbb{C}$ under the (inverse) stereographic projection. * Any circle that does pass the north pole corresponds to a line in $\mathbb{\tilde C}$ * *think of the line wrapping up at the point $\infty$* * Hence the **circline** is a geometric shape that is either a line or a circle in $\mathbb{\tilde C}$. * A circline is the set $\left\{ z: \left| \frac{z- \alpha}{z- \beta} \right| = \lambda\right\}$ * a line has $\lambda=1$, and it is the perpendicular bisector of the segment joining $\alpha$ and $\beta$. * a circle has $\lambda \ne 1$, and it is the Apollonius's circle. ### Möbius Transformations * **Mobius transformations** map circlines to circlines. They include rotation, dilation, translation, and inversion. * The first three transformations are obviously Mobius, and we can verify the last in the equation $|\frac{z-\alpha}{z-\beta}|= \lambda$. * In particular, the image has the equation $|\frac{z-f(\alpha)}{z- f(\beta)}|=\lambda$. * A Mobius transformation always have the form $w \to \frac{aw-b}{cw-d}$where $a,b,c,d \in \mathbb{C},\,ad-bc \ne 0$. * With this expression, we see that mobius transformations are bijective. ## Holomorphic Functions ### Differentiation in the Complex Plane * A function $f$ defined on an open subset $G \subseteq\mathbb{C}$ is differentiable at $z$ if $\lim_{ h \to 0 } \frac{f(z+h)-f(z)}{h}$exists, and the limit $L$ is its **derivative** has $\forall \epsilon>0,\, \exists \delta>0: h \in B(0,\delta) \Rightarrow \frac{f(z+h)-f(z)}{h} \in B(L, \epsilon)$ * This limit must hold in any direction: including (but not limited to) when $h$ approaches $0$ along the complex/real axes. * For example, $z \to z^2$ is differentiable at $z=0$, but $z \to\mathrm{Re}(z)$ is not differentiable any where (limits are different along the real and imaginary axes). * If $f$ has derivative $f'(z)$ at $z$, then$\begin{align*}f (z+h)=&f(z)+hf'(z)+h \epsilon(h),\\ \\ &\text{where } \epsilon(h)=\frac{f(x+h)-f(x)}{h}-f'(z)\to 0 \end{align*}$ * Hence $f(z+h)-f(z) \to 0$, $f$ continuous at $z$. * If the function $f(x+iy)=u(x,y)+iv(x,y)$ is differentiable at $z=x+iy$, it satisfy the **Cauchy-Riemann equations** at $z$: $u_{x}=v_{y};\, u_{y}=-v_{x}$ > Proof: writing the derivative as a limit along the real axis gives $f'=u_{x}+iv_{x}$, and $f'=-iu_{y}+v_{y}$ along the imaginary axis gves the second. Equating the two gives the Cauchy-Riemann equations. * Cauchy-Riemann equations are necessary but not sufficient for differentiability. They only account for the derivative along the real and imaginary axes, but not the other directions. * For example, the function $f=\mathbb{1}_{\mathrm{Re}(z)\ne0} \times \mathbb{1}_{\mathrm{Im}(z)\ne0}=\mathbb{1}_{\text{not on axis}}$ satisfies the equations at $z=0$ but is not differentiable there. ### Holomorphic Functions * A function that is differentiable over the entire $\mathbb{C}$ is **holomorphic**. * A function differentiable over an open subset $G \subseteq \mathbb{C}$ can be called holomorphic on $G$. * The differentiation rules (linearity, sum, product, chain) hold for functions holomorphic on relevant open sets. * Quotients of holomorphic functions are still holomorphic, provided the divisor is non-$0$ over the set. * Hence in particular, **rational functions** $p(x) / q(x)$, where $p,q$ are polynomials, are holomorphic over open sets where $q \ne 0$. * Identity theorems: on an open, connected subset $G \subseteq \mathbb{C}$, holomorphic $f$ is constant if any of the following is true: * $f'=0$ over $G$ * $|f|=c$ is constant * $f$ only has real values over $G$ > Proof: (1) holomorphism requires the Cauchy-Riemann equations to hold, and $f'=0$ along the real and imaginary axes in particular gives $u_{x}=v_{x}=u_{y}=v_{y}=0$. Applying the identity theorem to $u,v$ as single-variate real functions (given that one of $x$ and $y$ is fixed) gives $f$ constant over any horizontal/vertical segment in $G$. $G$ connected means that any two points can be joined by such segments, so the function is constant over $G$. > (2) Constant $c$ means $u^2+v^2=c$, so differentiate the equation wrt. $x$ and $y$ allows us to solve for $u_{x}=v_{x}=u_{y}=v_{y}=0$ in the Cauchy-Riemann equations. Again this means $f$ constant. > (3) Real-valued $f$ has $v=0$, so the Cauchy-Riemann equations again give $ $u_{x}=v_{x}=u_{y}=v_{y}=0$, $f$ constant. * As a result, any non-constant real function is not holomorphic: this incluse $|f|,\mathrm{Re}(f), \mathrm{Im}(f)$. ### Singularities * For a function $f$, a point is a **regular point** if $f$ is holomorphic there. * A point is a **singularity** if it is the limit point of regular points, but is not regular itself. * For example, $z=0$ is a singularity of $f(z)= z^{-1}$. * A singularity $a$ is an **isolated singularity** if $\exists \epsilon>0:$ $f$ is holomorphic on $B'(a, \epsilon)$. Otherwise, it is **non-isolated**. * that is, this singularity is an isolated point of the set of singularities. * Given function $f$ with Laurent expansion $\sum^{\infty}_{n=-\infty} c_{n}(z-a)^{n}$ an isolated singularity can be: * **Removable**, if $\forall n<0: c_{n}=0$ (so redefining the function at the singularity would make it converge to the Laurent/Taylor series and hence holomorphic). * A **pole** of order $k$, where $-k$ indexes the "last" non-$0$ coefficient: $c_{-k} \ne 0, c_{n<-k}=0$. * An **essential singularity** if otherwise (that is, $\not \exists k: \forall n<-k,c_{n}=0$). ### Poles * A pole of order $k$ of $f$ is also a zero of order $k$ of $f^{-1}$. * $f$ has a pole of order $k$ at $a \iff$ $\exists D \ne 0:\lim_{ z \to a }(z-a)^{k}f=D$. * Poles in products: if at $a$, $f$ has a pole of order $m$ and $g$ holomorphic near $a$, * $fg$ has pole of order $m-n$ if $g$ has a zero of order $n<m$. * $fg$ has a removable singularity if $n \ge m$. * $fg$ has a pole of order $m$ if $g(a) \ne 0$. * If $f,g$ has a pole of order $m,n$ resp. at $a$, then $fg$ has one of order $m+n$. * For example, $(z \sin(z))^{-1}$ has a pole of order $2$ at $0$, with one order from $z^{-1}$ and another from $\sin^{-1} (z)$. * If $f$ has a pole of order ### Meromorphic Functions * A function $f$ is **meromorphic** in $G$ open (in $\mathbb{C}$ or $\mathbb{\tilde{C}}$) if it is either holomorphic or has a pole at all $a \in G$. ### Residues and Cauchy's Residue Theorem * The **residue** of a function $f$ around $a$ is the coefficient $c_{-1}$ in its Laurent expansion around $a$. It is denoted $\text{res}(f;a)$. * Lemma around one pole: if $f$ has a pole of any order around $a$, and it is holomorphic on some $\bar{B}(a,r)$, then $\int _{\gamma(a,r)} f(z) \, dz=2\pi i \,\text{res}(f;a)$and by the deformation theorem, any positively oriented contour $\gamma$ with $f$ holomorphic in and on it. * **Cauchy's residue theorem**: given a positively-oriented contour $\gamma$, and function $f$ is holomorphic in and on it, except for the poles $a_{1}, \dots a_{n}$ (finitely many), then $\int _{\gamma} f(z) \, dz=2\pi i\sum^{n}_{k=1} \text{res}(f;a_{k})$so it is basically summing the effect around each pole. > Proof: let $f_{k}$ denote the principal part of $f$ (negative-powered terms in the Laurent expansion) around $a_{k}$. Then at any $b \ne a_{1 \dots n}$, all of $f_{1 \dots n}$ are holomorphic, so is $g=f- \sum_{k}f_{k}$. > Note that $g$ has removable singularities at $a_{1 \dots n}$, so redefining if necessary (which does not change the integral), $\int _{\gamma} g(z) \, dz=0$ since $g$ holomorphic in and on $\gamma$. So $\int _{\gamma} f \, dz= \sum_{k} \int _{\gamma} f_{k} \, dx=\sum_{k} \int _{\gamma(a_{k},r_{k})} f_{k} \, dx=2\pi i\sum^{n}_{k=1} \text{res}(f;a_{k})$where $f_{k}$ is holomorphic in and on $\gamma(a_{k},r_{k})$ except at $a_{k}$. ### Computing Residues * If $a$ is a simple pole of $f$, then $\text{res}(f;a)=\lim_{ z \to a }(z-a)f(z) $ * Furthermore, if $f=h / k$, where $h,k$ holomorphic around $a$, $k(a)=0$, $h(a),k'(a) \ne 0$, then $\text{res}(f;a)=h(a) / k'(a)$ > Proof: by AOL and the previous result, $\text{res}(f;a)=\lim_{ z \to a } (z-a)h/ k=h(a)\lim_{ z \to a }(z-a) / (k(a)-k(0))$This is $h(a) / k'(a)$. * The **Indentation Lemma** states that the integral on an indent around a simple pole is proportional to the length of the indent; in particular, with he indent being $\gamma(a,\epsilon):[\theta_{1}, \theta_{2}] \to \mathbb{C},\,0 \le \theta_{1} < \theta_{2} \le 2\pi$, there is $\lim_{ \epsilon \to 0 } \int _{\gamma(a,\epsilon)} f(z)\, dz \to i\, \text{Res}(f;a)(\theta_{1}-\theta_{2}) $which is proportional to the portion of the circle traversed. * this is only valid arounf a simple pole--a higher-ordered pole would blow up too quickly and make the integral infinite. > Proof: write $b=\text{Res}(f;a)$, then $\lim_{ z \to a }(z-a)f=b$. So $|(z-a)f-b| \to 0$. > Then noting that on $\gamma(a,\epsilon),z'(\theta)=\epsilon ie^{i\theta}=i(z-a)$, $\begin{align*} \bigg| \int_{\gamma(a,\epsilon)} f \, dz - ib(\theta_{1}-\theta_{2}) \bigg| &= \bigg|\int_{\theta_{1}}^{\theta_{2}} f(z(\theta))z'(\theta) - ib \, d\theta\bigg|\\ &=\bigg|i\int_{\theta_{1}}^{\theta_{2}} f(z)(z-a)-b \, d \theta \bigg|\\ &\le \int _{\theta_{1}}^{\theta_{2}} |(z-a)f-b| \, d \theta \to 0 \end{align*}$