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Cauchy Riemann & Taylor Expansion.

  1. Jul 8, 2008 #1
    Hi There. Was working on these and I think I managed to get most of them but still have a few niggling parts. I've managed to do questions 2,3,3Part2 and I've shown my working out so I'd be greatful if you could verify whether they are correct.
    Please could you also guide me on Q1 & 4.


    Q1: Suppose that f(z) = u(x,y) + iv(x,y) is holomorphic on a domain D, and that |f(z)| = constant throughout D. Show that f(z) = constant aswell.

    I realise this has something to do with the Cauchy Riemann equations but I can't quite apply it to this question. Any hints would be greatly appreciated.
    ----------------------------------------------------------------------------------------
    Q2:
    [itex]
    \begin{gathered}
    {\text{Using sum formula for cosine, write the taylor expansion of cos z centred at z}}_{\text{0}} {\text{, where z}}_0 {\text{ is a point in }}\mathbb{C}{\text{.}} \hfill \\
    {\text{Sum formula for cosine: }}\sum\limits_{n = 0}^\infty {( - 1)^n } \frac{{z^{2n} }}
    {{(2n)!}}. \hfill \\
    {\text{cos(z) : (z - z}}_{\text{0}} ){\text{ - }}\frac{{(z - z_0 )^2 }}
    {{2!}}{\text{ + }}\frac{{(z - z_0 )^4 }}
    {{4!}}{\text{ - }}\frac{{(z - z_0 )^6 }}
    {{6!}}............ \hfill \\
    {\text{Is this correct?}} \hfill \\
    \end{gathered}
    \]
    [/itex]

    ----------------------------------------------------------------------------------------
    Q3:

    [itex]
    \begin{gathered}
    {\text{Prove: }}\frac{d}
    {{dz}}\sin z = \cos z \hfill \\
    {\text{We know the definition of a derivative of a function f at z}} \in \mathbb{C}{\text{ is: f'(z) = }}\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{f(z + h) - f(z)}}
    {h}} \right]. \hfill \\
    {\text{Using the sine angle rule: sin(z + h) = sin(z)cos(h) + cos(z)sin(h)}}{\text{. So }}\frac{{\sin (z + h){\text{ }} - {\text{ }}\sin (z)}}
    {h}{\text{ }} = {\text{ }}\frac{{\sin (z)\cos (h){\text{ }} + {\text{ }}\cos (z)\sin (h){\text{ }} - {\text{ }}\sin (z)}}
    {h}{\text{ = }} \hfill \\
    \frac{{{\text{sin}}\left( {\text{z}} \right)\left( {{\text{cos}}\left( {\text{h}} \right){\text{ }} - {\text{ 1}}} \right){\text{ }} + {\text{ cos}}\left( {\text{z}} \right){\text{sin}}\left( {\text{h}} \right)}}
    {h}{\text{ = sin(z)}}\left[ {\frac{{\cos (h) - 1}}
    {h}} \right]{\text{ + cos(z)}}\left[ {\frac{{\sin (h)}}
    {h}} \right]. \hfill \\
    {\text{So now computing the derivative:}} \hfill \\
    \frac{d}
    {{dz}}\sin z{\text{ }} = {\text{ sin(z)}}\left[ {{\text{ }}\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos (h) - 1}}
    {h}} \right]} \right]{\text{ + cos(z)}}\left[ {{\text{ }}\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\sin (h)}}
    {h}} \right]} \right]{\text{ = sin(z)(0) + cos(z)(1) = 0 + cos(z) = cos(z)}}{\text{.}} \hfill \\
    {\text{Is this correct?}} \hfill \\
    \end{gathered} [/itex]

    ----------------------------------------------------------------------------------------
    Q3, Part 2:

    [itex]
    \begin{gathered}
    {\text{Prove: }}\frac{d}
    {{dz}}\cos z = - \sin z \hfill \\
    {\text{We know the definition of a derivative of a function f at z}}\varepsilon \mathbb{C}{\text{ is: f'(z) = }}\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{f(z + h) - f(z)}}
    {h}} \right]. \hfill \\
    {\text{Using the sine angle rule: cos(z + h) = cos(z)cos(h) - sin(z)sin(h)}}{\text{. So }}\frac{{\cos (z + h){\text{ }} - {\text{ }}\cos (z)}}
    {h}{\text{ }} = {\text{ }}\frac{{{\text{cos(z)cos(h) - sin(z)sin(h) }} - {\text{ }}\cos (z)}}
    {h}{\text{ = }} \hfill \\
    \frac{{{\text{cos}}\left( {\text{z}} \right)\left( {{\text{cos}}\left( {\text{h}} \right){\text{ }} - {\text{ 1}}} \right){\text{ }} - {\text{ sin}}\left( {\text{z}} \right){\text{sin}}\left( {\text{h}} \right)}}
    {h}{\text{ = cos(z)}}\left[ {\frac{{\cos (h) - 1}}
    {h}} \right]{\text{ - sin(z)}}\left[ {\frac{{\sin (h)}}
    {h}} \right]. \hfill \\
    {\text{So now computing the derivative:}} \hfill \\
    \frac{d}
    {{dz}}\cos z{\text{ }} = {\text{ cos(z)}}\left[ {{\text{ }}\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos (h) - 1}}
    {h}} \right]} \right]{\text{ - sin(z)}}\left[ {{\text{ }}\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\sin (h)}}
    {h}} \right]} \right]{\text{ = cos(z)(0) - sin(z)(1) = 0 - sin(z) = - sin(z)}}{\text{.}} \hfill \\
    {\text{Is this correct?}} \hfill \\
    \end{gathered}
    [/itex]

    ----------------------------------------------------------------------------------------
    Q4:

    Prove [itex]sin^2z + cos^2z = 1 [/itex] for all [itex]{\text{z}} \in \mathbb{C}.[/itex]


    No idea on this one. Please could you give me a hint.
     
  2. jcsd
  3. Jul 8, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    For Q1, if |f(z)| is constant then so is u^2+v^2. Take the partial derivatives of that with respect to x and y and apply Cauchy-Riemann. What do you conclude? For the other ones, it looks like you are trying to prove some trig type identities for sin and cos of a complex argument. So are you allowed to use these trig identities to prove them?? What is your definition of sin(z) and cos(z)? You should probably start with that. If you ARE allowed to use the identities you are using, that makes Q4 really easy. Use an identity on cos(z-z).
     
  4. Jul 8, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ??? That's not true. f(z)= z2 is holomorphic on the entire complex plane but is NOT constant. Are there other conditions on f? For example if f is holomorphic on the entire complex plane and bounded then it is a constant.

     
  5. Jul 8, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, there is another condition on f. |f(z)| is constant.
     
  6. Jul 9, 2008 #5
     
  7. Jul 9, 2008 #6
    QUESTION 4
    Here's my attempt for QUESTION 4:

    [itex]\begin{gathered}
    \sin \theta = \frac{{e^{i\theta } - e^{ - i\theta } }}
    {{2i}} \hfill \\
    \cos \theta = \frac{{e^{i\theta } + e^{ - i\theta } }}
    {2} \hfill \\
    \hfill \\
    \sin ^2 \theta = \left[ {\frac{{e^{i\theta } - e^{ - i\theta } }}
    {{2i}}} \right]^2 = \left( {\frac{1}
    {{2i}}\left( {e^{i\theta } - e^{ - i\theta } } \right)} \right)\left( {\frac{1}
    {{2i}}\left( {e^{i\theta } - e^{ - i\theta } } \right)} \right) \hfill \\
    = - \frac{1}
    {4}\left[ {e^{2i\theta } - 2 + e^{ - 2i\theta } } \right] \hfill \\
    \hfill \\
    \cos ^2 \theta = \left[ {\frac{{e^{i\theta } + e^{ - i\theta } }}
    {2}} \right]^2 = \left( {\frac{1}
    {2}\left( {e^{i\theta } + e^{ - i\theta } } \right)} \right)\left( {\frac{1}
    {2}\left( {e^{i\theta } + e^{ - i\theta } } \right)} \right) \hfill \\
    = \frac{1}
    {4}\left[ {e^{2i\theta } - 2 + e^{ - 2i\theta } } \right] \hfill \\
    \hfill \\
    \sin ^2 \theta + \cos ^2 \theta = \frac{1}
    {4}e^{2i\theta } + \frac{1}
    {2} + \frac{1}
    {4}e^{ - 2i\theta } - \frac{1}
    {4}e^{2i\theta } + \frac{1}
    {2} - \frac{1}
    {4}e^{ - 2i\theta } = \frac{1}
    {2} + \frac{1}
    {2} = 1 \hfill \\
    \end{gathered}
    \]
    [/itex]

    Seems good to me but I don't know whether that would be counted as a "rigorous" proof?
     
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