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Math Help - Finding the solutions with complex power

  1. #1
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    Finding the solutions with complex power

    \displaystyle\ 3e^{4xi}-12e^{2xi+x}+45e^{2xi+2x}+3=0
    I would like to find the solution of this equation, but i do not even know which is the 'leading' term, as in the one with the highest power of t. How would i compare \displaystyle\ t^{4i} to \displaystyle \ t^{2i+2} to atleast find which is the leading term? How could i go about solving this?
    Last edited by mr fantastic; December 9th 2010 at 11:56 AM.
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    Junior Member BariMutation's Avatar
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    It definitely belongs in pre-university, but that's not to say it's not tricky.

    First, you need to remember what i stands for; the square root of -1. This is an imaginary number, which means we can't deal with it without going into advanced stuff. You can, however, drastically simplify this problem. Remember that:

    3t^{4i} = (3t^4)^i

    Think you can go from here?
    Last edited by BariMutation; December 9th 2010 at 07:05 AM.
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  3. #3
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    3 isn't risen to the power of 4i, only t is. So with that being the case, i am guessing you ment \displaystyle 3(t^4)^i

    With that in mind, are you suggesting i substitute \displaystyle t^{2i}=uand then somehow treat it as a quadratic?The problem i find with that is what do \displaystyle t, t^2 and t^{-1} get substituted with?
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  4. #4
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    Quote Originally Posted by BariMutation View Post
    It definitely belongs in pre-university, but that's not to say it's not tricky. 3t^{4i} = (3t^4)^i
    I totally disagree with that quote.
    Any problem dealing with complex exponentiation is definitely a university level: perhaps number theory or analysis.

    Complex exponents do not behave as expected.
    For example, if each of u,~v,~\&~w is a complex number then [u^v]^w\not=u^{vw}.
    Only in the case that n\in \mathbb{Z} can we say that [u^v]^n\not=u^{nv}

    Thus as applied to this problem it can be said that t^{4i}=[t^i]^4.
    Here is a further compilation: t^i=\exp[i\log(t)] where \log(t)=\ln(|t|)+i\;\text{arg}(t).

    It is not clear if t is a complex variable or a real variable.
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  5. #5
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    Quote Originally Posted by Plato View Post
    I totally disagree with that quote.
    Any problem dealing with complex exponentiation is definitely a university level: perhaps number theory or analysis.

    Complex exponents do not behave as expected.
    For example, if each of u,~v,~\&~w is a complex number then [u^v]^w\not=u^{vw}.
    Only in the case that n\in \mathbb{Z} can we say that [u^v]^n\not=u^{nv}

    Thus as applied to this problem it can be said that t^{4i}=[t^i]^4.
    Here is a further compilation: t^i=\exp[i\log(t)] where \log(t)=\ln(|t|)+i\;\text{arg}(t).

    It is not clear if t is a complex variable or a real variable.
    I was wondering the same... I tried to make a substitution y = t^i, but I ended up with y and t together and this held me back from posting here. I'll be waiting for the solution to this question, once the OP comes to clarify the question, it seems really interesting!
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    Quote Originally Posted by Plato View Post
    I totally disagree with that quote.
    Here is a further compilation: t^i=\exp[i\log(t)] where \log(t)=\ln(|t|)+i\;\text{arg}(t).

    It is not clear if t is a complex variable or a real variable.
    I assume exp stands for exponential? I must also admit i have never used the arg(t) function.
    Perhaps i should clarify this:

    This is not a problem i found in a textbook, in fact it is an equation deriving from a function's intersection with the Ox axis.

    Background: I solved a differential equation and found the form of a function y(x). As i was curious, i attempted to draw the graph of that function, which was in exponential form and with complex powers. In order to draw the graph, i wanted to see where the function intersected the Oy and Ox axis. In an attempt to find where it intersected the Ox axis, i made it y(x)=0. I worked to try to simplify the equation in order to find the solutions and eventually reached what has been presented as the above equation.

    I should note that, in order to simplify and make things clearer for me, i substituted \displaystyle t=e^x. That is where the t comes from.

    With that being said, i expected it not be the average problem. The reason i put it in Pre-University was because it wasn't calculus or differentiation and had nothing to do with those, it was an equation and i did not find a more suitable place to post it. I was an am still curious to how this can be solved for the sheer pleasure of it
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  7. #7
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    Quote Originally Posted by kamykazee View Post
    I should note that, in order to simplify and make things clearer for me, i substituted \displaystyle t=e^x. That is where the t comes from.
    Now are you saying that t^{4i} is really e^{4xi}~?
    If so rewrite the whole problem as it should be.
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  8. #8
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    Here is the rewritten equation:

    \displaystyle\ 3e^{4xi}-12e^{x(2i+1)}+45e^{2x(i+1)}+3=0

    Could anyone aid me in how it could be solved?
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  9. #9
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    Quote Originally Posted by kamykazee View Post
    Here is the rewritten equation:
    \displaystyle\ 3e^{4xi}-12e^{x(2i+1)}+45e^{2x(i+1)}+3=0
    In the future please post the question not what you think we need to know.
    Two of us have wasted time on this.

    In general if x~\&~y are real numbers then e^{x+yi}=e^x\left(\cos(y)+i~\sin(y)\right)
    Example: e^{x(2i+1)}=e^x\left(\cos(2x)+i~\sin(2x)\right)
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  10. #10
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    Thank you for your answers. I'm sorry i was not explicit enough in the beginning and for posing this question in an inappropriate place. \displaystyle |x|
    Last edited by kamykazee; December 10th 2010 at 02:32 AM.
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