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An Algebraic Derivation of the Square Root of i

Date: 10/18/2003 at 18:16:16
From: Ethan
Subject: imaginary numbers

I was reading some of the questions and responses on imaginary 
numbers, and one about the square root of i interested me.  I don't 
know what sine and cosine are, and that makes it tough for me to 
understand the equations stated in the response.  If it is possible, 
can you explain the answer without using those two operations?

                      sqrt(2)     sqrt(2)
I know that sqrt(i) = -------  +  ------- i
                         2           2

but I'd like to understand how you come up with that.

Date: 10/29/2003 at 21:53:48
From: Doctor Justin
Subject: Re: imaginary numbers

Hi Ethan,

I do not know which math courses you have taken, but the following 
explanation uses algebra.  That way we can avoid the sine and cosine
that you mentioned.  Considering your interest in imaginary numbers,
perhaps you will be able to follow along easily.

The positive and negative square roots of i must be pure imaginary,
real, or complex.

Where x and y are real,      

  xi*xi = -x^2  so a pure imaginary squared yields a real

    y*y = y^2   so a real squared yields a real

The number i is nonreal, and since the real and pure imaginary
possibilities of its square root result in real numbers, the square 
root of i must be complex.

Therefore, the square root of i may be expressed as:

  (a + bi)^2 = 0 + i

where a and b are non-zero real numbers. 

Expand the binomial:

  a^2 + 2abi - b^2 = (0) + (1)i


  (a^2 - b^2) + (2ab)i = (0) + (1)i

Because for all complex numbers, if (a + bi) = (c + di), then a = c
and b = d, the following can be deduced:

  a^2 - b^2 = 0

  2ab = 1

Solve the second equation for b:

  b = 1/(2a)

Substitute into the first equation:

  a^2 - (1/(2a))^2 = 0


  a^2 - (1/4a^2) = 0

Multiply by 4a^2:
  4a^4 - 1 = 0

  (2a^2 + 1)(2a^2 - 1) = 0

Because (2a^2 + 1) has imaginary roots, it can not be considered in 
the solution for a.  This is because, by definition, a, b, c, and d 
must each be real in the step we used earlier.


  2a^2 - 1 = 0

Solve for a:

  a = sqrt(2)/2


  b = 1/(2a)

Subsitute and solve for b:

  b = 1/[2*sqrt(2)/2] = 1/sqrt(2) = sqrt(2)/2


  (a + bi)^2 = i


  sqrt(i) = +- (a + bi)

  sqrt(i) = +- (sqrt(2)/2 + (sqrt(2)/2)i)

Factor, and thus:

  sqrt(i) = +- (sqrt(2)/2)(1 + i)

I hope that helps.  If you are still confused, please write back.

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

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