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Thread: Complex Function properties

  1. #1 Complex Function properties 
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    If I want to find some actual meaning from an equation containing complex numbers, can I do this:

    x(t) = (4 + 3i)*(exp(-2t) - exp(5it))

    Re[x(t)] = Re[4 + 3i] * ( exp(-2t) - Re[exp(5it)] )

    where I just take the real portion of each complex number, and drop the imaginary parts?

    I feel like I'm missing something - someone let me know. Thanks!


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  3. #2 Re: Complex Function properties 
    . DrRocket's Avatar
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    Quote Originally Posted by ConiunctoErgoSum
    If I want to find some actual meaning from an equation containing complex numbers, can I do this:

    x(t) = (4 + 3i)*(exp(-2t) - exp(5it))

    Re[x(t)] = Re[4 + 3i] * ( exp(-2t) - Re[exp(5it)] )

    where I just take the real portion of each complex number, and drop the imaginary parts?

    I feel like I'm missing something - someone let me know. Thanks!
    Yep you are missing quite a bit. The real part of a product is not the product of the real parts.


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  4. #3  
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    so... what, then, is the real part of a product?
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  5. #4  
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    Quote Originally Posted by ConiunctoErgoSum
    so... what, then, is the real part of a product?
    Work it out for yourself. It is extremely easy.
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  6. #5  
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    Let and . Find .
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  7. #6  
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    Quote Originally Posted by Ellatha
    Let and . Find .
    You have the right idea.

    But your notation suggests two functions f and g of some variable x. There is no "x" on the right side of the "=" sign, nor is one needed.

    All one needs to do is work out Re(a+bi)(c+di) vs ac.
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  8. #7  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by Ellatha
    Let and . Find .
    You have the right idea.

    But your notation suggests two functions f and g of some variable x. There is no "x" on the right side of the "=" sign, nor is one needed.

    All one needs to do is work out Re(a+bi)(c+di) vs ac.
    I agree that the notation I used was rather sloppy. If one prefers, it could be viewed as and . Constant functions, i.e., are similar.

    I also agree that the answer to this question is extremelly simple:




    Now let and . Therefore, multiplying the binomials derives the binomial , where is the real part and C is the real coefficient of the imaginary number .
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