1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where is the error in this argument

  1. Feb 16, 2013 #1
    It is possible to do this and it is correct:

    \log \left[\sqrt{i}\right] = \log\left\{\exp\left[\frac{i}{2}\left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = \frac{i}{2}\left(\frac{\pi}{2} + 2\pi n\right) = i\left(\frac{\pi}{4} + \pi n\right)


    \log \left[i^2 \right] = \log\left\{\exp\left[2i \left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = 2i \left(\frac{\pi}{2} + 2\pi n\right) = i\left(\pi + 4\pi n\right)


    ## \log \left[i^2 \right] = \log \left[-1 \right] = i\left(\pi + 2\pi n\right) \ for \ k \in \mathbb{Z}## which is the correct argument.
  2. jcsd
  3. Feb 16, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure what your argument really is. Sure (sqrt(i))^4=(i)^2=(-1). (sqrt(-i))^4 is also (-1). So?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Where is the error in this argument
  1. Epsilon argument (Replies: 3)

  2. Deside the argument (Replies: 7)