Very simple complex powers problem

[Aziza]

Write the following in x+iy form: (-1)^i

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...

 

[Muphrid]

For integer k, sure they are. Any full turn around the unit circle doesn't change the value.

 

[Aziza]

For integer k, sure they are. Any full turn around the unit circle doesn't change the value.

but e^i2πk and e^2πk 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

 

[Muphrid]

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.

 

[Aziza]

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...

 

[Curious3141]

Write the following in x+iy form: (-1)^i

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.

 

[SammyS]

Write the following in x+iy form: (-1)^i

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...

(Those π's are hard to read !)

Sure enough.

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

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

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