# Abstract Nonsense

## Complex Differentiable and Holmorphic Functions (Pt. II)

Point of Post: This is a continuation of this post.

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Complex Differentiablility

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Let $\Omega\subseteq\mathbb{C}$ be open (we are obviously endowing $\mathbb{C}\approx\mathbb{R}^2$ with the usual topology). We say that a function $f:\Omega\to\mathbb{C}$ is differentiable at $z_0\in\Omega$ if there exists a $\mathbb{C}$-linear function $A:\mathbb{C}\to\mathbb{C}$ such that the limit

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$\displaystyle \lim_{z\to z_0}\frac{f(z)-f(z_0)-A(z-z_0)}{z-z_0}=0$

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We call $A$ the complex total derivative of $f$ at $z_0$ and denote it $\widetilde{D}_f((z_0)$. Of course, this definition is pure pedantry since any $\mathbb{C}$-linear function $\mathbb{C}\to\mathbb{C}$ is of the form $z\mapsto \alpha z$ for some complex number $\alpha$. Thus, we know that $\widetilde{D}_f(z_0)$, if it exists, is secretly just multiplication by some constant. We denote this constant $f'(z_0)$ and call it the derivative of $f$ at $z_0$. We see then that $f$ being differentiable at $z_0$ is equivalent to the statement that

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$\displaystyle \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$

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exists, and when it does, it is equal to $f'(z_0)$.

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If $f$ is differentiable at every point of $\Omega$ we say it’s differentiable on $\Omega$. Note then that we get a function $f':\Omega\to\mathbb{C}$ defined by $z_0\mapsto f'(z_0)$. We say that $f$ is continuously differentiable if $f'$ is continuous on $\Omega$. We say that $f$ is twice differentiable if $f'$ is differentiable on $\Omega$, and we define $n$-times differentiable similarly.

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The first key theorem about differentiable functions is the following:

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Theorem: Let $\Omega\subseteq\mathbb{C}$ be open. Then, if $f:\Omega\to\mathbb{C}$ is such that $f$ is complex differentiable at $z_0$ then it is real differentiable at $z_0=x_0+iy_0=(x_0,y_0)$ and $\widetilde{D}_f(z)=D_f(z)$ (i.e. the complex total derivative is equal to the real total derivative).

Proof: Identifying $\mathbb{C}\cong\mathbb{R}^2$ as $\mathbb{R}$-vector spaces we see that $\widetilde{D}_f(z)$ is naturally an $\mathbb{R}$-map $\mathbb{R}^2\to\mathbb{R}^2$ which evidently satisfies

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$\displaystyle \lim_{(x,y)\to (x_0,y_0)}\frac{\|f(x,y)-f(x_0,y_0)-\widetilde{D}_f(x-x_0,y-y_0)\|}{\|(x,y)-(x_0,y_0)\|}=0$

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from where the conclusion follows by the definition of being real differentiable and being a total derivative. $\blacksquare$

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This allows us to conclude about the complex derivative all the things that were true for real differentiable functions $\mathbb{R}^2\to\mathbb{R}^2$. For example:

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Corollary(Chain Rule): Let $\Omega,\Omega'\subseteq\mathbb{C}$ be open and suppose that $f:\Omega\to\Omega'$ and $g:\Omega'\to\mathbb{C}$ are functions such that $f$ is differentiable at $z_0\in\Omega$ and $g$ is differentiable at $f(z_0)\in\Omega$. Then, $g\circ f:\Omega\to\mathbb{C}$ is differentiable at $z_0$ and $(g\circ f)'(z_0)=g'(f(z_0))\cdot f'(z_0)$.

Proof: By the chain rule for total derivatives and the fact that $\widetilde{D}=D$ (complex total derivative equals total derivative) we know that $\widetilde{D}_{g\circ f}(z_0)=\widetilde{D}_g(f(z_0))\circ \widetilde{D}_f(z_0)$, but this is clearly just the linear map $z\mapsto g'(f(z_0))f'(z_0)$ from where the conclusion follows. $\blacksquare$

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Using the same idea we also can conclude the following:

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Corollary: Let $\Omega\subseteq\mathbb{C}$ be open and $f,g:\Omega\to\mathbb{C}$. Suppose that $f,g$ are differentiable at $z_0\in\Omega$ then:

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\begin{aligned}&\mathbf{(1)}\quad f+g\text{ is differentiable at }z_0\text{ and }(f+g)'(z_0)=f'(z_0)+g'(z_0)\\ &\mathbf{(2)}\quad fg\text{ is differentiable at }z_0\text{ and }(fg)'(z_0)=f'(z_0)g(z_0)+f(z_0)g'(z_0)\\ \displaystyle &\mathbf{(3)}\quad \frac{f}{g}\text{ is differentiable at }z_0\text{ and }\left(\frac{f}{g}\right)'(z_0)=\frac{g(z_0)f'(z_0)-f(z_0)g'(z_0)}{g(z_0)^2}\end{aligned}

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where we must assume that $g(z_0)\ne 0$ for the last one.

Proof: The first and last one we have proven before (here and here respectively) and the second can be proved thinking of multiplication as a bilinear map and applying this result. $\blacksquare$

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We also know this fact:

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Theorem: Let $\Omega\subseteq\mathbb{C}$ be open and connected. Then, if $f'=g'$ as functions on $\Omega$ then $f=g+c$ for some constant $c$. In particular, if $f'=0$ then $f$ is constant.

Proof: Apply the case for real functions. $\blacksquare$

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Since we know that real differentiable functions are continuous we may also conclude this:

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Theorem: Let $\Omega\subseteq\mathbb{C}$ be open. If $f:\Omega\to\mathbb{C}$ is differentiable at $z_0\in\Omega$ then it is continuous at $z_0$.

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References:

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[1] Greene, Robert Everist, and Steven George Krantz. Function Theory of One Complex Variable. Providence: American Mathematical Society, 2006. Print.

[2] Conway, John B. Functions of One Complex Variable I. New York: Springer-Verlag, 1978. Print.

[3] Rudin, W. Real and Complex Analysis. New York,NY: McGraw-Hill, 1988. Print.

[4]  Ahlfors, Lars V. Complex Analysis; an Introduction to the Theory of Analytic Functions of One Complex Variable. New York: McGraw-Hill, 1966. Print.