- #1

Aziza

- 190

- 1

my solution:

(-1)

^{i}= (e

^{i(π+2πk)})

^{i}= e

^{-(π+2πk)}

However, my book just states the answer as e

^{-(π)}...but these are not the same...

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- Thread starter Aziza
- Start date

- #1

Aziza

- 190

- 1

my solution:

(-1)

However, my book just states the answer as e

- #2

Muphrid

- 834

- 2

- #3

Aziza

- 190

- 1

but e

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

- 834

- 2

- #5

Aziza

- 190

- 1

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

- 2,858

- 88

my solution:

(-1)^{i}= (e^{i(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

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,693

- 1,273

(Those π's are hard to read !)

my solution:

(-1)^{i}= (e^{i(π+2πk)})^{i}= e^{-(π+2πk)}

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

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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