The Square Root of i

Date: 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/