Very simple complex powers problem

  • Thread starter Aziza
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  • #1
Aziza
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Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) ...but these are not the same...
 

Answers and Replies

  • #2
Muphrid
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For integer [itex]k[/itex], sure they are. Any full turn around the unit circle doesn't change the value.
 
  • #3
Aziza
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For integer [itex]k[/itex], sure they are. Any full turn around the unit circle doesn't change the value.

but ei2πk and e2πk are different...the first one equals 1 but the second does not...

i mean look just plug in k=1 and k=2...you get e^-3 and e^-5, which are not the same
 
  • #4
Muphrid
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Oh, okay, I see what's going on. Yeah, in this case, you may be more correct than the book is. Someone with more recent experience than I in complex analysis may know something more specific to this, though.
 
  • #5
Aziza
190
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Oh, okay, I see what's going on. Yeah, in this case, you may be more correct than the book is. Someone with more recent experience than I in complex analysis may know something more specific to this, though.

oh alright. idk my book is doing this same mistake for each and every related problem. so i was thinking that i must be doing something wrong. but maybe they are just giving the answer when k=0...although they never specify this...
 
  • #6
Curious3141
Homework Helper
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Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) ...but these are not the same...

Your answer is more complete than the book's. Exponentiation to a complex power can give a multivalued result. The book just gave the principal value, which, if you're asked for a single answer, is the usual value we take.
 
  • #7
SammyS
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Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) ...but these are not the same...
(Those π's are hard to read !)

Sure enough.

Give WolframAlpha, [itex]\displaystyle \left(e^{\pi} \right)^{1/i}[/itex] or [itex]\displaystyle \left(e^{-3\pi} \right)^{1/i}\,,[/itex] and it gives -1.

Of course, [itex]\displaystyle e^{-3\pi}\approx 0.0000806995175703\,,[/itex] [itex]\displaystyle e^{-\pi}\approx 0.04321391826377226\,,[/itex] [itex]\displaystyle e^{\pi}\approx 23.140692632779263\,,[/itex]

So it looks like [itex]\displaystyle (-1)^i=e^{(\text{odd integer multiple of }\pi)}\ .[/itex]
 

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