The Square Root of iDate: 05/25/97 at 21:04:01 From: Leif Subject: The square root of i What is the square root of i? Date: 05/26/97 at 07:58:07 From: Doctor Anthony Subject: Re: The square root of i We have i = cos(pi/2) + i.sin(pi/2) sqrt(i) = (cos(pi/2) + i.sin(pi/2))^(1/2) By DeMoivre's theorem: = cos(pi/4) + i.sin(pi/4) 1 + i = ---------- sqrt(2) To learn more about DeMoivre and his theorem, look at: http://mathforum.org/library/drmath/view/53975.html -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 12/08/2003 at 07:58:07 From: Doctor Schwa Subject: Re: The square root of i Good question! When I was in high school and confronted with this same problem, it seemed obvious to me that the answer must be "j". Just like like we needed to invent a new number "i" to be the square root of -1, it seemed like we'd need yet another new kind of number to be the square root of "i", and so on forever. The amazing thing is that you don't: once you have "i", any equations with addition, multiplication, exponents and so on (in short, any polynomial)' can be solved without inventing any new types of numbers! An equation like x+3 = 2 makes you invent negatives, x*3 = 2 makes you invent fractions, x*x = 2 makes you invent irrationals, x*x = -2 makes you invent imaginaries... but then you're done! So, once I knew it was possible, and that the answer had to be some complex number (a+bi), the question is, how do you find out values of a and b that will make (a+bi)^2 = i? Well, squaring out the left side gives a^2 + 2ab * i - b^2 = i, and the only way for that to work is if the real number part is 0, a^2 - b^2 = 0, and the imaginary part is 1*i, 2ab = 1. Since a^2 = b^2, a = b or -b ... but since 2ab = 1, a and b must be both positive or both negative, so a = b. Then since 2ab = 1, and a = b, 2aa = 1, so a^2 = 1/2, and a = b = sqrt(1/2) or a = b = -sqrt(1/2)! Does that make sense? -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/