hey all, i got this problem

find all the solutions z^(i)=1

so i wrote it out as e^(i(ln(1)+i@)= 1

@= theta

and i end up getting e^(-@)=1

is that right?

Printable View

- Feb 28th 2010, 07:09 PMwowsomemez^(i)=1
hey all, i got this problem

find all the solutions z^(i)=1

so i wrote it out as e^(i(ln(1)+i@)= 1

@= theta

and i end up getting e^(-@)=1

is that right? - Mar 1st 2010, 03:35 AMtonio
- Mar 1st 2010, 04:53 AMdrumist

with - Mar 1st 2010, 05:02 AMLaurent
This question also appeared in this thread.

To complete the answer I gave there (since Tonio's answer is incorrect): is equivalent to (up to this point, the values of the left-hand sides depend on the determination of the logarithm), which is equivalent to for some , and thus to , . Hence the solutions are the real numbers for every ; it is worth noting that since the answer is real, the choice of the determination of the logarithm is irrelevant (and you can forget about it if you don't know what I mean).

Edit: drumist was quicker than me, but I leave my post because of the remark on the definition of logarithm - Mar 1st 2010, 11:57 AMwowsomeme
ok i get that, but how do u knoe arg is 0?