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Loser66

  • one year ago

Check my stuff, please. z = i w = -1 +i zw = -1-i

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  1. anonymous
    • one year ago
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    Ill help Brb

  2. Loser66
    • one year ago
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    arg z = pi/2 arg w = 3pi/4 arg (zw) = 5pi/4, what is wrong with them?

  3. anonymous
    • one year ago
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    Nevermind I cant help But I give Medal And Fan Sry

  4. Loser66
    • one year ago
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    @Empty

  5. Empty
    • one year ago
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    Nothing is wrong with them, arg(z)+arg(w)=arg(zw) in general.

  6. Loser66
    • one year ago
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    what is log (zw) ?

  7. Loser66
    • one year ago
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    I have to prove log (zw) \(\neq \) log z + log w because of their argument from both sides are not the same. But I don't see it :(

  8. Empty
    • one year ago
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    Depends, it's multivalued since \[e^{i \theta} = e^{i(\theta + 2 \pi)}\] we can do this an integer amount of times + or - so we have: Every complex number has a polar form: \[z=re^{i(\theta + 2 \pi n)}\] \[\log(z) =\log (re^{i(\theta + 2 \pi n)}) = \log (r) + i(\theta + 2 \pi n)\]

  9. anonymous
    • one year ago
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    your'e correct

  10. Empty
    • one year ago
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    http://math.stackexchange.com/a/927129/207119

  11. Empty
    • one year ago
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    You're trying to prove a false statement as far as I can tell.

  12. Loser66
    • one year ago
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    (zw) = -1-i, hence its argument is 5pi/4

  13. Loser66
    • one year ago
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    and this argument is = arg z + arg w, right?

  14. Empty
    • one year ago
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    \[\frac{\pi}{2}=\frac{2 \pi}{4}\] \[\frac{2 \pi}{4} + \frac{3 \pi}{4} = \frac{5 \pi}{4} \]

  15. Loser66
    • one year ago
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    So, for those numbers, z and w. The statement is true. right? log (zw) = log z + log w

  16. Loser66
    • one year ago
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    Hence, to prove the statement is wrong, I have to pick other z, w, right?

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