MTH2003 Complex Analysis Pre-Lecture 4

• Recording 1.
• Recording 2.
• Lecture Notes.
• Lecture: MTH2003 Complex Analysis Lecture 4.
• Practical (MTH2003 Complex Analysis Practical 2).

Paths in the Complex Plane

A path $\gamma$ connects two points, $\gamma (a)$ and $\gamma (b)$, in $\mathbb{C}$ with a continuous curve:

$$\gamma :[a,b]\to \mathbb{C},\text{where }[a,b]\subset \mathbb{R}\text{(a continuous parametric function)}$$

These paths can be combined (${\gamma }_1+{\gamma }_2$, although this addition is only notation) to produce a single path, as long as the end of the first path is the beginning of the second path. This can be done by finding the path functions (e.g. ${\gamma }_1(t)=\sqrt{2}{e}^{it}$ and ${\gamma }_2=(1-t)u+tv$), including their intervals. When integrating, you can then add them.

Example (Combining Paths): Let ${\gamma }_1(t)=2e^{it}$ for $t\in [0,\pi /2]$ (a quarter-circle from $2$ to $2i$), and ${\gamma }_2(t)=(1-t)2i+t(2)$ for $t\in [0,1]$ (a line segment from $2i$ to $2$). The combined path ${\gamma }_1+{\gamma }_2$ is defined by reparameterizing ${\gamma }_2$ over $[\pi /2,\pi /2+1]$ and concatenating the intervals.

A closed path satisfies $\gamma (a)=\gamma (b)$ (i.e. it ends where it began). Example: $\gamma (t)=r{e}^{it}$ for $t\in [0,2\pi ]$ (a circle of radius $r$).

Every path also has an opposite path $-\gamma :[a,b]\to \mathbb{C}$ such that $-\gamma (t):=\gamma (a+b-t)$ - essentially just the reverse. Other notation is $\tilde{\gamma }$. Example: If $\gamma (t)=t$ for $t\in [0,1]$, then $-\gamma (t)=1-t$.

A subset $S\subset \mathbb{C}$ is called path-connected if for every pair of points $p,q\in S$ there is a continuous path $\gamma :[a,b]\to \mathbb{C}$ with $\gamma (a)=p$ and $\gamma (b)=q$. Open disks $D=\left\{z\in \mathbb{C}:|z|<a\right\}$ are path-connected.

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Differentiation

A function $f:D\to \mathbb{C}$ is differentiable at $z_0\in D$ if the limit:

$$f^{\prime }(z_0)=\underset{h\to 0}{\lim }\frac{f(z_0+h)-f(z_0)}{h}$$

exists independently of the path $h$ takes to approach $0$. If $f^{\prime }(z_0)$ exists, then all higher-order derivatives exist, and $f$ is analytic (holomorphic) on $D$. Analytic functions satisfy:

• Linearity: $(af+bg{)}^{\prime }=a{f}^{\prime }+b{g}^{\prime }$.
• Product Rule: $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$.
• Quotient Rule: ${\left(\frac{f}{g}\right)}^{\prime }=\frac{f^{\prime }g-f{g}^{\prime }}{g^2}$.
• Chain Rule: $(f\circ g{)}^{\prime }(z)=f^{\prime }(g(z))g^{\prime }(z)$.

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Partial Derivatives & Jacobian Matrix

For $f(z)=u(x,y)+iv(x,y)$, the Jacobian matrix is:

$$Df(x,y)=\left[\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right].$$

If $f$ is differentiable, $Df(x,y)$ must represent a complex linear map, equivalent to multiplication by $f^{\prime }(z)$. This enforces the Cauchy-Riemann equations:

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Cauchy-Riemann Theorem

Theorem: Let $f(z)=u(x,y)+iv(x,y)$ be defined on an open set $D\subset \mathbb{C}$. Then $f$ is differentiable at $z_0=x_0+i{y}_0$ if and only if:

1. $u_x,u_y,v_x,v_y$ exist and are continuous near $z_0$.
2. The Cauchy-Riemann equations hold at $z_0$:

$$u_x=v_y,u_y=-v_x.$$

Proof (Sketch):

• ($\Rightarrow$) If $f$ is differentiable, compute $f^{\prime }(z_0)$ by approaching $z_0$ along the real axis ($h=\Delta x$) and imaginary axis ($h=i\Delta y$). Equating results gives $u_x=v_y$ and $v_x=-u_y$.
• ($\Leftarrow$) If $u,v$ have continuous partials satisfying C-R, use the total differential to show the complex limit exists.

Remark: The Jacobian matrix of $f$ satisfies $Df=\left[\begin{array}{cc}{u}_x& u_y\\ v_x& v_y\end{array}\right]$. When C-R holds, this matrix becomes $\left[\begin{array}{cc}{u}_x& -v_x\\ v_x& u_x\end{array}\right]$, equivalent to multiplication by the complex number $u_x+i{v}_x=f^{\prime }(z)$.

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Example (Verification for $f(z)=e^z$): Write $e^z=e^x\cos y+i{e}^x\sin y$, so $u=e^x\cos y$, $v=e^x\sin y$. Compute:

• $u_x=e^x\cos y=v_y$,
• $u_y=-e^x\sin y=-v_x$.

Thus, C-R equations hold everywhere, so $e^z$ is entire (analytic on $\mathbb{C}$).

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Example (Finding Analytic Functions): Find all analytic $f(z)$ with $\text{Re}(f)=u(x,y)=x^2-y^2$.

1. Compute partial derivatives: $u_x=2x$, $u_y=-2y$.
2. By C-R: $v_x=-u_y=2y$, $v_y=u_x=2x$.
3. Integrate $v_x$ with respect to $x$: $v=2xy+g(y)$.
4. Differentiate $v$ with respect to $y$: $v_y=2x+g^{\prime }(y)=2x⟹g^{\prime }(y)=0⟹g(y)=C$.
5. Thus, $v=2xy+C$, and $f(z)=(x^2-y^2)+i(2xy+C)=z^2+iC$. Since $C$ is a real constant, the general solution is $f(z)=z^2+iC$ for $C\in \mathbb{R}$.

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Key Takeaways:

• Paths in $\mathbb{C}$ are parametrized curves; combinations require matching endpoints.
• Analytic functions are infinitely differentiable and satisfy C-R equations.
• C-R equations link partial derivatives of real and imaginary parts, ensuring complex differentiability.