Loser66
  • Loser66
Check my stuff, please. z = i w = -1 +i zw = -1-i
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Ill help Brb
Loser66
  • Loser66
arg z = pi/2 arg w = 3pi/4 arg (zw) = 5pi/4, what is wrong with them?
anonymous
  • anonymous
Nevermind I cant help But I give Medal And Fan Sry

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Loser66
  • Loser66
@Empty
Empty
  • Empty
Nothing is wrong with them, arg(z)+arg(w)=arg(zw) in general.
Loser66
  • Loser66
what is log (zw) ?
Loser66
  • Loser66
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 :(
Empty
  • Empty
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: \[z=re^{i(\theta + 2 \pi n)}\] \[\log(z) =\log (re^{i(\theta + 2 \pi n)}) = \log (r) + i(\theta + 2 \pi n)\]
anonymous
  • anonymous
your'e correct
Empty
  • Empty
http://math.stackexchange.com/a/927129/207119
Empty
  • Empty
You're trying to prove a false statement as far as I can tell.
Loser66
  • Loser66
(zw) = -1-i, hence its argument is 5pi/4
Loser66
  • Loser66
and this argument is = arg z + arg w, right?
Empty
  • Empty
\[\frac{\pi}{2}=\frac{2 \pi}{4}\] \[\frac{2 \pi}{4} + \frac{3 \pi}{4} = \frac{5 \pi}{4} \]
Loser66
  • Loser66
So, for those numbers, z and w. The statement is true. right? log (zw) = log z + log w
Loser66
  • Loser66
Hence, to prove the statement is wrong, I have to pick other z, w, right?

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