Complex analysis problem

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  • #1
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Homework Statement



Consider the equation
(z-1)^23 = z^23

Show that all solutions lie on the line Re(z)=1/2
How many solutions are there


Homework Equations





The Attempt at a Solution


Really have no idea. I figured polar form might be helpful somehow so I converted it and got
(r(cos (theta) + isin (theta)) - 1)^23 - r^23(cos (23*theta) + isin (23*theta))=0
I'm getting hung up on the first term. I'm not really sure what I can do with it.
 

Answers and Replies

  • #2
Hootenanny
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It would probably be better to begin with to write the equation in exponential notation and notice that [itex]-1 = \exp\{i\pi\}[/itex].
 
  • #3
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Alright, I've got
(r23ei[itex]\theta[/itex]-ei[itex]\pi[/itex])23-r23e23i[itex]\theta[/itex]

I don't recall any way to combine the two exponents. Expanding those brackets are going to be a mess, I think, and probably unnecessary since I'm not trying to find the solutions exactly.
 
Last edited:
  • #4
HallsofIvy
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It would make more sense if you had an equal sign in there!
 
  • #5
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It would, wouldn't it! That equation should have an =0 at the end. My apologies.
 
  • #6
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How about you just muscle-through it with z's first:

[tex]\frac{z-1}{z}=1^{1/23}[/tex]

Brave huh?

[tex]z=\frac{1}{1-e^{2n\pi i/23}}[/tex]

Now, even though it's messy, can't you now convert everything to sines and cosines, rationalize the denominator, and just split-out the real part of that and hopefully, it'll be 1/2?

Hope that's not too much help guys.
 
  • #7
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Thanks, jackmell. I didn't think of doing that. Once I did, the rest of the problem proved to be simple.
 
  • #8
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jackmell could you explain that step just a bit more?

did you just rearrange so that we had unity on the right side of the equation, and then take the 23rd root of all the values?

not quite sure how you got to the next step either?
maybe brandy can explain?

man I'm bad at complex analysis!
 
  • #9
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That's exactly what he did. It's a nice way to eliminate that annoying exponent that I didn't think of until I saw him do it.

The next part is just rearranging what he had to get it in the form z=

It isn't an easy subject. Unfortunately, the prerequisite subject for complex analysis at my uni didn't cover the assumed knowledge so I've got a bit of catching up to do.
 
  • #10
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Just thought a bit more on that solution... Isn't it undefined everywhere since e2in[itex]\pi[/itex]=1 and therefore the denominator is 0?
 
  • #11
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Ok, lemme' see if I can get out of this jam . . .

We started with [tex](z-1)^{23}=z^{23}[/tex]

Ok, if you expanded that out, you'd get not a 23rd degree polynomial but a 22 degree polynomial right since you'd remove z^23. Ok, so that's 22 solutions.

Now, during the algebra, we got to:

[tex]1-1^{23}=1/z[/tex]

Usually, we then just take the reciprocal but actually we're dividing by 1-1^23 in one step then multiplying by z in the next. When we divide by 1-1^23, we're assuming that's not zero. And 1-1^23 is zero only when the argument is zero right? Because:

[tex]1^{23}=e^{i/23(\theta+2n\pi)}[/tex]

and only one of those roots is one and that's when theta=0. So there we go, that's 22.
 
  • #12
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Ah, thanks guys. I confused myself with the exponential for a bit.
 

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