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Cauchy.Riemann equations

  1. Jun 2, 2014 #1
    Hello.
    Here I attached the question and what I have done so far. I have used the cauchy.riemann equations to check if the functions are analytical or not. The first one turns out to be analytical, but I don't know how to find its derivative function and how do I express it as a z function?
     

    Attached Files:

  2. jcsd
  3. Jun 2, 2014 #2

    CAF123

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    Hint: Rewrite the first equation as $$f(x,y) = e^x x (\cos y + i \sin y) + e^x y (i\cos y - \sin y)$$

    Now do some further manipulations and simplify, so as to be left with something purely in terms of ##z = x+iy##.
     
  4. Jun 2, 2014 #3
    But what about this formula? I got this from a textbook. f'(z0) here is the derive function.
     

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  5. Jun 2, 2014 #4

    CAF123

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    Once you rewrite ##f(x,y)## as defined in the last post entirely in terms of ##z##, then computing df/dz is arguably easier.

    Use Euler's formula when simplifying ##f(x,y)##.

    Edit: I just saw your attachment. That is the definition of the derivative. But generally speaking there are techniques you should use such as product rule which are easier to deal with for more complicated functions.
     
  6. Jun 2, 2014 #5
    So the one I just attached is not a formula but actually just a definition of the derivative?
     
  7. Jun 2, 2014 #6

    CAF123

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    It can be used in practice too, but even with the most simplest of functions you'd better off using the power rule, product rule etc..

    E.g assuming you are familiar with the real version, let ##f(x) = x^2##. By power rule, ##df/dx = 2x##. Done.

    By the definition of the derivative, $$f'(x) = \lim_{h \rightarrow 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h \rightarrow 0} \frac{2hx + h^2}{h} = 2x.
    $$
    Then try it with ##f(x) = \sqrt{x}## and already more manipulations needed.
     
  8. Jun 2, 2014 #7
    Is this what you mean? Did I do it correctly?
     

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  9. Jun 2, 2014 #8

    CAF123

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    Close, but ##-1/i = i##, so you messed up a sign. Other than that, looks fine. Now use the fact that ##z = x+iy## to obtain a function all in terms of z.

    Note also by simply factoring out ##i## from the second term you get ##ie^x y(\cos y + i\sin y) = ie^x ye^{iy}##, so no need to mess about with more definitions.
     
  10. Jun 2, 2014 #9
    Now is this correct?
    So, next how do I obtain a function?
     

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  11. Jun 2, 2014 #10

    CAF123

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    Yes, it is correct. But you can simplify further. ##e^a e^b = \dots## and use that z = x+iy.
     
  12. Jun 2, 2014 #11
    You mean like this? Is that the final answer?
     

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  13. Jun 2, 2014 #12

    CAF123

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    One last simplification you can make. Make sure you have only z in your expression (no appearances of x or y). Then are you after df/dz?
     
  14. Jun 2, 2014 #13
    Aha! I think I got it. Is this right?
     

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  15. Jun 2, 2014 #14

    CAF123

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    Yeah, it is good now.
     
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