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Math Help - Complex numbers CONTRADICTION?

  1. #1
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    Complex numbers CONTRADICTION?

    hey guys
    1. I proved that z^-1=conj z, converting into mod-arg form and by doing de moivres theorem so this must therefore be true
    1/(2+3i)=2-3i

    2. yet when i substituted z for an actual complex number eg. z=2+3i
    and then did the realisation the answer is different
    1/(2+3i) X (2-3i)/(2-3i)=(2-3i)/13

    3. and i learned of a complex number rule where
    z^-1=conj z/(mod z)^2
    and this rule is consistent with what got ... which was (2-3i)/13

    so now the question finally is why are there 2 answers that seem right?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aonin View Post
    hey guys
    1. I proved that z^-1=conj z, converting into mod-arg form and by doing de moivres theorem so this must therefore be true
    1/(2+3i)=2-3i

    2. yet when i substituted z for an actual complex number eg. z=2+3i
    and then did the realisation the answer is different
    1/(2+3i) X (2-3i)/(2-3i)=(2-3i)/13

    3. and i learned of a complex number rule where
    z^-1=conj z/(mod z)^2
    and this rule is consistent with what got ... which was (2-3i)/13

    so now the question finally is why are there 2 answers that seem right?
    It's definitely untrue that z^{-1}=\bar{z} since as you noted for z\ne 0 you have that z^{-1}=\frac{\bar{z}}{|z|^2} from where it quickly follows that \bar{z}=z^{-1}\implies z\in\mathbb{S}^1 and the converse can be checked quite readily.
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  3. #3
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    but it is true z^-1=conj z through demoivres theorem though?. what is the symbol that looks like an E, sorry im new here
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    What were you asked to do originally?

    You have all these work based on something wrong.

    Are you asked to prove or disprove if z^{-1}=\bar{z} \ \mbox{?} If so, an example of why it doesn't suffices which you have. If not, what are you trying to do?
    Last edited by dwsmith; December 26th 2010 at 04:46 PM. Reason: Verbage
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  5. #5
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    There are several mistakes in your OP.
    Consider this fact: z^{ - 1}  = \dfrac{1}{z} = \dfrac{{\overline z }}{{\left| z \right|^2 }}.

    What does that do to what you posted?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aonin View Post
    but it is true z^-1=conj z through demoivres theorem though?. what is the symbol that looks like an E, sorry im new here
    I don't see how DeMoivre's theorem: \displaystyle (\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin  (n\theta), implies this. Aha! I bet I know where you got messed up! You can get from DeMoivre's theorem that \left(\cos(\theta)+i\sin(\theta)\right)^{-1}=\cos\left(-1\theta\right)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta)=\overline{\cos(\theta)+i\sin(\theta)  } and you had a fuzzy idea that every number can be represented as something 'like' \cos(\theta)+i\sin(\theta) but you forgot that it's actually r(\cos(\theta)+i\sin(\theta)) where r=|z|. Thus, this proves once again that the two coincides when 1=r=|z|. As for your question \mathbb{S}^1 is just the unit circle |z|=1.
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  7. #7
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    there was a question regarding graphing 1/z and conj z and they were the same line
    then i realised that through demoivre's theorem that the conj and 1/z could be proved to be equal

    however when i realised the denominator in 1/z where eg. z=2+3i the result is (2+3i)/13 which is consistent to the rule z^-1=conj z/(mod z)^2

    so why are there 2 answers by using 2 methods (demoivre's theorem and realising the denominator), or am i misunderstanding something?
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  8. #8
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    yep i get what you mean now. i just forgot about the modulus when doing (rcistheta)^-1=r^-1 X cis-theta. thanks for clearing that up
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  9. #9
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    You can't graph Complex Numbers like Real Numbers since you are dealing with 4 dimensions.

    w=f(\bar{z}) \ \ \{w\in\mathbb{C}:w\neq 0\}

    \displaystyle\mbox{Circle:} \ w=f\left(\frac{1}{z}\right) \ \ \left\{w\in\mathbb{C}:w\neq 0 \ |w|=\left|\frac{1}{r}\right|\right\}

    \displaystyle x=k: \ \  w=f\left(\frac{1}{z}\right) \ \ \left\{w\in\mathbb{C}:w\neq 0 \ \left|w-\frac{1}{2k}\right|=\left|\frac{1}{2k}\right|\righ  t\}

    \displaystyle y=k: \ \ w=f\left(\frac{1}{z}\right) \ \ \left\{w\in\mathbb{C}:w\neq 0 \ \left|w+\frac{1}{2k}\mathbf{i}\right|=\left|\frac{  1}{2k}\right|\right\}

    Extended Complex Plane includes w=0.
    Last edited by dwsmith; December 26th 2010 at 06:24 PM. Reason: Cleaning up appearance
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  10. #10
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    Quote Originally Posted by aonin View Post
    there was a question regarding graphing 1/z and conj z and they were the same line
    then i realised that through demoivre's theorem that the conj and 1/z could be proved to be equal

    however when i realised the denominator in 1/z where eg. z=2+3i the result is (2+3i)/13 which is consistent to the rule z^-1=conj z/(mod z)^2

    so why are there 2 answers by using 2 methods (demoivre's theorem and realising the denominator), or am i misunderstanding something?
    Again we have a situation where, if the original question had been posted rather than some misleading/incorrect interpretation of it, a lot of time and effort would have been saved ....
    Last edited by mr fantastic; December 31st 2010 at 05:16 PM. Reason: Test edit.
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