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Complex Analysis Properties Question 2

  1. Jun 16, 2015 #1

    RJLiberator

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    The problem states, Show that:
    a) |e^(i*theta)| = 1.
    Now, the definition of e^(i*theta) makes this
    |cos(theta)+isin(theta)|
    If we choose any theta then this should be equal to 1.

    What can help me prove this? If I choose, say, pi/6 then it simplifies to |(sqrt(3))/2+i/2)| which doesn't seem to equal 1.




    b) BAR(e^(i*theta)) = e^(-i*theta)

    Here's what I think. The bar e^(i*theta) means that the definition is cos(theta)-isin(theta) and this can be rewritten as cos(-theta)+i*sin(-theta) which can be inputted back into the definition to see that e^(-i*theta) is correct.
     
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  3. Jun 16, 2015 #2

    Dick

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    Uh, ##|x+iy|=\sqrt{x^2+y^2}##, yes? So?
     
  4. Jun 16, 2015 #3

    RJLiberator

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    Ok, so when we have |cos(theta)+isin(theta)| this can be represented as sqrt(cos^2(theta)+sin^2(theta)) and this is clearly equal to 1, always.

    Ah, that is simply beautiful.
    I appreciate your guidance here. Swiftly helped me here.
     
  5. Jun 20, 2015 #4

    micromass

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    It does.
     
  6. Jun 20, 2015 #5

    WWGD

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    Also, I am not sure what the starting point is, what you can assume, but ## cos\theta+ isin\theta ## is a parametrization for a point in the unit circle.
     
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