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FP2 complex exponents watch

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    I am having a thick day today, i am really stuck with this question.

    Find all the solutions of, and plot them on an argand diagram.

    e^z = (1 -√3.j)/(2e^4)

    Im not worried about the argand diagram, i just need to know how you would solve this. Thanks for the help!
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    Re-arrange it:
    e^(z+4) = (1/2) - (sqrt(3)/2)j

    And use:
    e^(z+4) = cos(z+4) + j sin(z+4)

    Now equate real and imaginary parts, and remember that there are infinitely many solutions for each equation.
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    I am still a little confused, i see what you have done but i dont know the last step, does this mean i need to evaluate:

    1/2 = cos(z+4)
    (sqrt(3)/2)j = jsin(z+4)

    I dont get how you can equate the real and imaginary parts, because z surely still could contain a real and imaginary part, no?
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    Ah, an oversight on my part.

    We could go back to the original equation, and notice that:
    e^(z+4) = (1/2) - (sqrt(3)/2)j = cos(pi/3 + 2npi) - j sin(pi/3 + 2npi) = e^(j(-pi/3) + 2npi)
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    Thank you!!
    i have the answer i needed now, however i think your post is slightly incorrect. would i be right in saying that this is not right:
    = e^(j(-pi/3) + 2npi)
    and that it should be:
    = e^j((-pi/3) + 2npi)

    If that is right i can rearange it perfectly to get the answer i should. Thanks.
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    Yes, that's right.

    2 slips in 1 thread! That clearly means one thing: I shouldn't be doing any maths at 8am. :p:
 
 
 
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Updated: April 9, 2006

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