Some changes have been made to LiveJournal, and we hope you enjoy them! As we continue to improve the site on a daily basis to make your experience here better and faster, we would greatly appreciate your feedback about these changes. Please let us know what we can do for you!

See a bug? Let us know! Here you can also share your thoughts and ideas about updates to LiveJournal

Your request has been filed. You can track the progress of your request at:

If you have any other questions or comments, you can add them to that request at any time.

Help us make LiveJournal better! Answer a few questions to share your experiences using LiveJournal with us. It won't take more than 10 minutes!

Take a survey
alsterellieMay 20 2009, 22:37:21 UTC 6 years ago

nekokazeMay 20 2009, 22:43:02 UTC 6 years ago

roadriverrailMay 20 2009, 22:49:16 UTC 6 years ago

nekokazeMay 20 2009, 23:04:09 UTC 6 years ago

There is no value k that will do so, real or complex. The limit of e^x as x approaches negative infinity is 0. That's the best you can do, since negative infinity isn't really a "value" per se, it only has meaning in the limit statement.

nekokazeMay 20 2009, 23:11:52 UTC 6 years ago

Say complex w = xi + y for real x and y.

e^w = e^(xi+y) = e^(xi) * e^y

We know e^y cannot be 0, thus if e^w =0 it must be e^(xi) that is equal to 0.

But by Euler's formula:

e^(xi) = i * sin(x) + cos(x)

0 = i * sin(x) + cos(x) => i * sin(x) = -cos(x)

But neither sin(x) nor cos(x) may have an imaginary value for real x, thus one side of the equation is imaginary and the other is not. Therefore the equation has no solution.

Thus if e^x = 0 lacks a solution for real x, it also lacks one for complex x.

roadriverrailMay 20 2009, 23:19:52 UTC 6 years ago

This sort of thing doesn't come up often in, say, operating systems development, so I haven't touched these topics since college.

korean_guy_01May 20 2009, 22:53:06 UTC 6 years ago

nekokazeMay 20 2009, 23:05:24 UTC 6 years ago

roadriverrailMay 20 2009, 22:45:44 UTC 6 years ago

Unless there's an imaginary number out there that suffices.

makingthematrixMay 21 2009, 06:39:30 UTC 6 years ago

cr4kMay 21 2009, 18:38:04 UTC 6 years ago

triprcMay 21 2009, 19:55:54 UTC 6 years ago

cr4kMay 21 2009, 21:28:24 UTC 6 years ago

greedyalgorithmMay 21 2009, 20:52:36 UTC 6 years ago

jtootfwrote inalgorithms: ←n-th order statistics of the immutable datatriprcwrote inalgorithms: →(No Subject)