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Square Root of i

Date: 12/06/2001 at 20:38:22
From: Morgan McBee
Subject: Square root of i

What is the square root of i (square root of the square root of 
negative one)?

On my calculator, I get square root of two divided plus i times the 
square root of two divided by two, but I do not understand why.

Date: 12/06/2001 at 21:04:45
From: Doctor Tom
Subject: Re: Square root of i

Hi Morgan,

You can test that your calculator is right: simply multiply that 
number by itself and you'll get i.

The reason is not obvious.  See Imaginary Exponents and Euler's 
Equation from the Dr. Math FAQ:   

- Doctor Tom, The Math Forum   

Date: 12/06/2001 at 22:09:49
From: Doctor Pete
Subject: Re: Square root of i


To try to answer your question in a satisfying way, I'll introduce a 
definition for the imaginary unit i that does not involve square roots 
of negative numbers. We set up a system of addition and multiplication 
of ordered pairs (a,b), where a and b are real numbers, in the 
following manner:

For any two pairs (a,b) and (c,d), we have

     (a,b) + (c,d) = (a+c, b+d),
     (a,b) * (c,d) = (ac-bd, ad+bc).

Then we see that numbers of the form (a,0) correspond to real numbers, 

     (a,0) + (c,0) = (a+c, 0)
     (a,0) * (c,0) = (ac - 0, 0) = (ac, 0).

That is, there is a one-to-one correspondence between the ordered pair 
(a,0) and the real number a. So we might as well say (a,0) = a.  
Furthermore, we note that

     (0,1) * (0,1) = (0 - 1, 0 + 0) = (-1, 0) = -1.

So the ordered pair (0,1) has the property that its square is -1.  
Therefore, we define

     i = (0,1),

and we can say for all real numbers a, b, that

     a + bi = (a,b),


     a + bi = (a,0) + (b,0)*(0,1) = (a,0) + (b*0 - 0*1, b*1 + 0*0)
            = (a,0) + (0,b)
            = (a,b).

So instead of thinking of complex numbers as things like 3 + 4i, we 
can think of them as ordered pairs like (3,4), which follow the rules 
of addition and multiplication that we specified earlier.

Now, the "square root of i" under this new definition is an ordered 
pair (x,y), with x, y real numbers, such that

     (x,y)^2 = (x,y)*(x,y) = (0,1).

But we use the definition of multiplication to find

     (x,y)^2 = (x^2 - y^2, 2xy) = (0,1).

If we match up the corresponding parts of the ordered pairs, we get 
the system of equations

     x^2 - y^2 = 0,
           2xy = 1.

If we consider the first equation, we have

     x^2 - y^2 = (x+y)(x-y) = 0,

which means x+y = 0, or x-y = 0.  In the first case, x = -y, and 
substituting this into the second equation gives

     2(-y)y = 1,

which has no real solution for y.  In the second case, x = y, and 
substituting into the second equation then gives

     2y^2 = 1,
     y = {1/Sqrt[2], -1/Sqrt[2]}.

Hence, we have found two solutions, namely

     (x,y) = {(1/Sqrt[2], 1/Sqrt[2]), (-1/Sqrt[2],-1/Sqrt[2])},

and if we want to write this in the more conventional x + yi notation,

    x + yi = {(1+i)/Sqrt[2], (-1-i)/Sqrt[2]}.

I hope you don't think my explanation was too complicated. It is a 
little unconventional, because most people would explain it by talking 
about the polar form of complex numbers, and De Moivre's Theorem.

- Doctor Pete, The Math Forum   
Associated Topics:
High School Imaginary/Complex Numbers

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