# Math 416 Final Review Story after Cauchy's theorem ## Chapter 7: Cauchy's Theorem ### Existence of harmonic conjugate Suppose $f=u+iv$ is holomorphic on a domain $U\subset\mathbb{C}$. Then $u=\Re f$ is harmonic on $U$. That is $\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$. Moreover, there exists $g\in O(U)$ such that $g$ is unique up to an additive imaginary constant. > Example: > > Find a harmonic conjugate of $u(x,y)=\log|\frac{z}{z-1}|$ > > Note that $\log(\frac{z}{z-1})=\log \left|\frac{z}{z-1}\right|+i(\arg(z)-\arg(z-1))$ is harmonic on $\mathbb{C}\setminus\{1\}$. > > So the harmonic conjugate of $u(x,y)=\log|\frac{z}{z-1}|$ is $v(x,y)=\arg(z)-\arg(z-1)+C$ where $C$ is a constant. > > Note that the harmonic conjugate may exist locally but not globally. (e.g. $u(x,y)=\log|z(z-1)|$ has a local harmonic conjugate $i(\arg(z)+\arg(z-1)+C)$ but this is not a well defined function since $\arg(z)+\arg(z-1)$ is not single-valued.) ### Corollary for harmonic functions #### Theorem 7.19 Harmonic function are infinitely differentiable. #### Theorem 7.20 Mean value property of harmonic functions. Let $u$ be harmonic on an open set of $\Omega$, then $$u(z_0)=\frac{1}{2\pi}\int_0^{2\pi} u(z_0+re^{i\theta}) d\theta$$ for any $z_0\in\Omega$ and $r>0$ such that $D(z_0,r)\subset\Omega$. #### Theorem 7.21 Identity theorem for harmonic functions. Let $u$ be harmonic on a domain $\Omega$. If $u=0$ on some open set $G\subset\Omega$, then $u\equiv 0$ on $\Omega$. #### Theorem 7.22 Maximum principle for harmonic functions. Let $u$ be a non-constant real-valued harmonic function on a domain $\Omega$. Then $|u|$ does not attain a maximum value in $\Omega$. ## Chapter 8: Laurent Series and Isolated Singularities ### Laurent Series Laurent series is a generalization of Taylor series. Laurent series is a power series of the form $$f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$$ where $$ a_k=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{(z-z_0)^{k+1}}dz $$ The series converges on an annulus $R_1<|z-z_0| Example: > > $f(z)=\frac{e^z-1}{z^2}$ has a removable singularity at $z=0$. > > The Laurent series of $f$ at $z=0$ can be found using the Taylor series of $e^z-1$ at $z=0$. > > $$e^z-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$ > > So the Laurent series of $f$ at $z=0$ is > > $$f(z)=\frac{1}{z^2}+\frac{1}{z}+\sum_{n=0}^{\infty}\frac{z^n}{n!}$$ > > The principle part is zero, so $z=0$ is a removable singularity. #### Poles