Complex Number Problem #1

By KS Teng

I’ve come across many questions from students regarding complex numbers, and here is one thats quite fun to deal with!

\sqrt{-1}^{\sqrt{-1}}

Now this is really interesting, but can be solved with a bit of manipulation. We all know this is akin to i^i

We know that \sqrt{-1}=e^{i\frac{\pi}{2}}

\sqrt{-1}^{\sqrt{-1}}

= e^{i\frac{\pi}{2}\sqrt{-1}}

= e^{i\frac{\pi}{2}i}

= e^{i^2\frac{\pi}{2}}

= e^{-\frac{\pi}{2}}

Now our end result is a real number!

Like mentioned with regards to Euler’s Identity, it is really very amazing how complex numbers actually work!

Comments
    pingbacks / trackbacks

    Leave a Comment

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt
    0

    Start typing and press Enter to search

    What is a Markov Process?University Mathematics
    2011 A-level H2 Mathematics (9740) Paper 2 Question 8 Suggested SolutionsJC Mathematics