Holomorphy and Calculus

fact:

Useful facts: \begin{align*} f' = {\frac{\partial f}{\partial z}\,} = {1\over i}{\frac{\partial f}{\partial y}\,} = {\frac{\partial f}{\partial x}\,} = {\frac{\partial u}{\partial x}\,} + i {\frac{\partial v}{\partial x}\,} .\end{align*}

Complex Calculus

fact:

Various differentials: \begin{align*} dz &= dx + i~dy \\ d\overline{z} &= dx - i~dy \\ \\ f_z &= f_x = f_y / i .\end{align*}

Integral of a complex exponential: \begin{align*} \int_{0}^{2 \pi} e^{i \ell x} d x &=\left\{\begin{array}{ll} {2 \pi} & {\ell=0} \\ {0} & \text{else} \end{array}\right. .\end{align*}

\begin{align*} \int_{\gamma} f(z) \,dz= \int_0^{2\pi } f(Re^{i\theta}) \, iRe^{i\theta} \,d\theta .\end{align*}

warnings:

f(z) = \sin(z), \cos(z) are unbounded on {\mathbf{C}}! An easy way to see this: they are nonconstant and entire, thus unbounded by Liouville.

example (Square root is not holomorphic):

You can show f(z) = \sqrt{z} is not holomorphic by showing its integral over S^1 is nonzero. This is a direct computation: \begin{align*} \int_{S^1} z^{1/2} \,dz &= \int_0^{2\pi} (e^{i\theta})^{1/2} ie^{i\theta} \,d\theta\\ &= i \int_0^{2\pi} e^{i3\theta \over 2}\,d\theta\\ &= i \qty{2\over 3i} e^{i3\theta \over 2}\Big|_{0}^{2\pi} \\ &= {2\over 3}\qty{e^{3\pi i - 1}} \\ &= -{4\over 3} .\end{align*}

Note an issue: a different parameterization yields a different (still nonzero) number \begin{align*} \cdots &= \int_{-\pi}^{\pi} (e^{i\theta})^{1/2} ie^{i\theta} \,d\theta\\ &= {2\over 3}\qty{ e^{3\pi i \over 2} - e^{-3\pi i \over 2}} \\ &= -{4i\over 3} .\end{align*} This is these are paths that don’t lift to closed loops on the Riemann surface defined by z\mapsto z^2.

remark (Solving real integrals with complex calculus):

The function e^z is entire, so “usual” calculus using it works. For example, \begin{align*} \int e^{3x}\cos(2x) \,dx &= \Re \int e^{3z}e^{2iz}\,dz\\ &= \Re \int e^{(3+2i)z} \,dz\\ &= \Re {e^{(3+2i)z} \over 3+2i} + C \\ &= e^{2x \over 13}\qty{\cos(2x) + i\sin(2x)} .\end{align*}

Holomorphy

remark:

Let f:{\mathbf{C}}\to {\mathbf{C}} satisfy CR and consider it as a map f:{\mathbf{R}}^2\to{\mathbf{R}}^2, writing f(x+iy) = u(x, y) + iv(x, y), the Jacobian is \begin{align*} J = u_xv_y -v_x u_y = u_x^2 + v_x^2 .\end{align*} OTOH, \begin{align*} {\left\lvert {f'(z)} \right\rvert}^2 = {\left\lvert {u_x + iv_y} \right\rvert}^2 = J .\end{align*}

definition (Complex differentiable / holomorphic /entire):

A function f: {\mathbf{C}}\to {\mathbf{C}} is complex differentiable or holomorphic at z_0 iff the following limit exists: \begin{align*} \lim_{h\to 0} { f(z_0 + h) - f(h) \over h } .\end{align*} A function that is holomorphic on {\mathbf{C}} is said to be entire.

Equivalently, there exists an \alpha\in {\mathbf{C}} such that \begin{align*} f(z_0+h) - f(z_0) = \alpha h + R(h) && R(h) \overset{h\to 0}\longrightarrow 0 .\end{align*} In this case, \alpha = f'(z_0).

example (Holomorphic vs non-holomorphic):

    
    
      
        
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          f(z) \coloneqq{\left\lvert {z} \right\rvert} is not holomorphic.
        
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          f(z) \coloneqq\arg{z} is not holomorphic.
        
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          f(z) \coloneqq\Re{z} is not holomorphic.
        
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          f(z) \coloneqq\Im{z} is not holomorphic.
        
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          f(z) = {1\over z} is holomorphic on {\mathbf{C}}\setminus\left\{{0}\right\} but not holomorphic on {\mathbf{C}}
        
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          f(z) = \overline{z} is not holomorphic, but is real differentiable: 

\begin{align*}
{f(z_0 + h) - f(z_0) \over h } = {\overline{z_0} + \overline{h} - \overline{z_0} \over h} = {\overline{h} \over h} = {re^{-i\theta} \over re^{i\theta}} = e^{-2i\theta} \overset{h\to 0}\longrightarrow e^{-2i\theta}
,\end{align*}
 which is a complex number that depends on \theta and is thus not a single value.
        
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definition (Real (multivariate) differentiable):
    

      A function F: {\mathbf{R}}^n\to {\mathbf{R}}^m is real-differentiable at \mathbf{p} iff there exists a linear transformation A such that 

\begin{align*}
{ {\left\lVert { F(\mathbf{p} + \mathbf{h}) - F(\mathbf{p}) - A(\mathbf{h}) } \right\rVert} \over {\left\lVert { \mathbf{h} } \right\rVert} } \overset{{\left\lVert {\mathbf{h}} \right\rVert}\to 0}\longrightarrow 0
.\end{align*}
 Rewriting, 

\begin{align*}
{\left\lVert { F(\mathbf{p} + \mathbf{h}) - F(\mathbf{p})  - A(\mathbf{h}) } \right\rVert} = {\left\lVert { \mathbf{h} } \right\rVert} {\left\lVert { R(\mathbf{h}) } \right\rVert}
&& {\left\lVert {R(\mathbf{h}) } \right\rVert}\overset{{\left\lVert {\mathbf{h} } \right\rVert} \to 0}\longrightarrow 0
.\end{align*}

    

  
    

      Equivalently, 

\begin{align*}
F(\mathbf{p} + \mathbf{h}) - F(\mathbf{p}) = A(\mathbf{h}) + {\left\lVert {\mathbf{h}} \right\rVert} R(\mathbf{h}) && {\left\lVert {R(\mathbf{h}) } \right\rVert}\overset{{\left\lVert {\mathbf{h} } \right\rVert} \to 0}\longrightarrow 0
.\end{align*}

    

  
    

      Or in a slightly more useful form, 

\begin{align*}
F(\mathbf{p} + \mathbf{h}) = F(\mathbf{p}) + A(\mathbf{h}) + R(\mathbf{h}) && R\in o( {\left\lVert {\mathbf{h}} \right\rVert}), \text{ i.e. }
{ {\left\lVert { R(\mathbf{h}) } \right\rVert} \over  {\left\lVert {\mathbf{h}} \right\rVert}} \overset{\mathbf{h}\to 0}\longrightarrow 0
.\end{align*}

    

  
proposition (Holomorphic implies continuous.):
    

      f is holomorphic at z_0 iff there exists an a\in {\mathbf{C}} such that 

\begin{align*}  
f(z_0 + h) - f(z_0) - ah = h \psi(h), \quad \psi(h) \overset{h\to 0}\to 0
.\end{align*}
 In this case, a = f'(z_0).
    

  

    

      #todo proof
    

  
Wirtinger Calculus
remark:
    

      Some properties:
    

  
    
    
      
        
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          \overline{{ \overline{{\partial}}}f(z)} = {\partial}\,\overline{f}(z), or {\partial}f^*(z) = ({ \overline{{\partial}}}f(z))^*
    
      
      
        
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          E.g. d(cz) = c\,dz+ 0\,d\overline{z} and d(c\overline{z}) = 0\,dz+ c\,d\overline{z}
        
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          E.g. {\partial}{\left\lvert {z} \right\rvert}^2 = {\partial}z\overline{z} = \overline{z} and { \overline{{\partial}}}{\left\lvert {z} \right\rvert}^2 = z.
    
        
      
        
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          So d({\left\lvert {z} \right\rvert}^2) = \overline{z}\,dz+ z\,d\overline{z}
        
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          E.g. {\partial}\exp\qty{ - {\left\lvert {z} \right\rvert}^2 } = {\partial}\exp\qty{-z\overline{z}} = e^{-{\left\lvert {z} \right\rvert}^2}\cdot {\partial}(z\overline{z}) = \overline{z} e^{-{\left\lvert {z} \right\rvert}^2}.
        
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Exercises
exercise (?):
    

      Show that a real-valued holomorphic function is constant.
    

  
    

      #complex/exercise/completed
    

  
solution:
    

      Use that {\mathbf{R}} is not open in {\mathbf{C}} and apply the open mapping theorem to conclude: f({\mathbf{C}}) must be open in {\mathbf{C}} if f is holomorphic and nonconstant.
    

  
exercise (?):
    

      Show that if fg \equiv 0 on a domain \Omega then either f\equiv 0 or g\equiv 0.
    

  
    

      #complex/exercise/completed
    

  
solution:
    

      If f\not\equiv 0, there is a point z_0 where f(z_0)\neq 0, and thus a neighborhood U\ni z_0 where f is nonvanishing. This forces g\equiv 0 on U, however U is a set with a limit point, so g\equiv 0 on \Omega by the identity principle.
    

  
exercise (?):
    

      Find all entire functions f such that f(x) = e^x on {\mathbf{R}}.
    

  
    

      #complex/exercise/completed
    

  
solution:
    

      The function g(z) \coloneqq f(z) - e^z is entire and identically zero on {\mathbf{R}}, which contains a limit point. So g(z) \equiv 0 on {\mathbf{C}}, meaning f(z) = e^z is the only such function.
    

  

  


            

              
              
            

            
  

    
      
        
        #todo
        
      
    
      
        
        #complex/exercise/completed