Is It True That j is Equal to the Square Root of -1 ?

Rick LyonsSeptember 16, 20136 comments

A few days ago, on the YouTube.com web site, I watched an interesting video concerning complex numbers and the j operator. The video's author claimed that the statement "j is equal to the square root of negative one" is incorrect. What he said was:

He justified his claim by going through the following exercise, starting with:

Based on the algebraic identity:

the author rewrites Eq. (1) as:

If we assume

Eq. (3) can be rewritten as:

The video's author reaches his goal by stating, because j2 = -1 we can write:

Arriving at Eq. (6) the video's author asks, "What's going on here?" He continues by explaining that because 1 ≠ -1, the definition in Eq. (4) must be an "erroneous definition." Thus, he instructs, we cannot say "j is equal to the square root of negative one."

I disagree, and here's why. The impossible result of 1 = -1 in Eq. (6) is not caused by the Eq. (4) definition, the error is caused by Eq. (2)! The fact is, the identity in Eq. (2) is not, in general, correct when dealing with negative arguments. Specifically, Eq. (2) is not correct when a and b are both negative, as they are in Eq. (1). So it is the above Eq. (3) that is not valid, and this causes the impossible result in Eq. (6).

As it turns out, we must be very careful when performing square roots involving negative numbers. (Appendix A gives an interesting example of reckless use of such square root operations.) That's because when dealing with negative numbers some of our time-honored algebraic root identities can lead to incorrect results. For example, the seemingly obvious equality

is not valid when argument a is a negative number. (Appendix B illustrates other time-honored algebraic identities, involving roots, that are not always valid.)

The author of the YouTube video ends his presentation by suggesting that, rather than ever writing Eq. (4), we should always write:

as the definition of the j operator. Equation (7) merely states that j is a number whose square is -1, and that equation will keep us out of mathematical trouble [1]. I wasn't able to understand the last 30 seconds of the video, but it seems to me that if we take the square root of both sides of Eq. (7) we obtain:

giving us the ambiguities of:

and we must choose whichever expression in Eq. (9) that gives us sensible results in our algebraic activities. So I agree with the video's author that Eq. (7) should be used to define the j operator because squaring both sides of the two expressions in Eq. (9) always produces Eq. (7).

So tonight at the dinner table, you may want to formally announce to your spouse and children, "Family, ...from now on we will not write j equals the square root of minus one. We will write j squared equals minus one."

[1] http://www.youtube.com/watch?v=5iCoBU0o86Q

Appendix-A [Careless Square Root Example]
The following is an example of silly results obtained if we’re careless when performing square roots.

Theorem: 4 = 5


16 - 36 = 25 - 45
42 - 9·4 = 52 - 9·5
42 - 9·4 + 81/4 = 52 - 9·5 + 81/4
(4 - 9/2)2 = (5 - 9/2)2
4 - 9/2 = 5 - 9/2
4 = 5

It's the sloppy square root operation that led to the invalid result of 4 = 5. In truth, the tricky author of the above proof should have written:

(4 - 9/2)2 = (5 - 9/2)2
±(4 - 9/2) = ±(5 - 9/2)
±(-1/2) = ±(1/2).

Appendix-B [Laws of Roots - When are they valid?]

In the following table, a and b are scalars and variables p and q are positive integers

[ - ]
Comment by harishmSeptember 17, 2013
Hi Richard,

I have a small doubt at Eq. (8) in above post, is it necessary to apply +/- sign on both sides ...?
[ - ]
Comment by Rick LyonsSeptember 17, 2013
Hello Harishm,
In my urge to be thorough I put the +/- signs in front of the 'j'. But you are correct, ...that is not necessary. I've requested web master Stephane Boucher to modify Equations (8) and (9). Thanks for pointing that out to me Harishm.

Sept. 18, 2013
[ - ]
Comment by russhJuly 14, 2017

Hi Rick,

The problem is with equations 1,2, and 3. You must include angles when working with complex numbers.

1= sqrt(1)= sqrt{(-1)(-1)'}= sqrt(-1)sqrt(-1)'=j(-j)=1.

The number 1 has an angle of 0 degrees in the complex world so 1 equals the product of -1 with the complex conjugate of -1 indicated as (-1)' in order for the angles to sum to zero. The complex conjugate of j is -j.

Equation 2 should be written:

sqrt{(aAngA)(bAngB)}=sqrt{a}Ang(A/2)sqrt{ b }Ang(B/2)
=sqrt{ab}Ang( A/2+B/2 )

By the way j is defined as +sqrt(-1).

Cheers Russ

Sorry about the math. It looked good in the editor but the math ml did not stick.

[ - ]
Comment by dsss27September 22, 2013
Hamilton solved quaternions via a similar identity:

i^2 = j^2 = k^2 = ijk = ?1
[ - ]
Comment by lxgSeptember 19, 2013
Hello, Richard,

The ambiguity still exists even using j^2=-1 to define j, because the j is clearly defined in the complex plane and the square operation can eliminate the difference between j and -j. I think it's due to that '1' is a pure real number, but 'j' is a more general complex number, therefore, maybe it's proper to define j in complex domain directly.
[ - ]
Comment by amirkhan3January 4, 2015
The correction for +/-(4-9/2) =+/-(5-9/2) is +/-(4-9/2) =-/+(5-9/2);
4-9/2 = -(5-9/2) = -1/2;
-(4-9/2) = (5-9/2) = 1/2;
By overlooking you see 4=5 but it is not.

To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.

Please login (on the right) if you already have an account on this platform.

Otherwise, please use this form to register (free) an join one of the largest online community for Electrical/Embedded/DSP/FPGA/ML engineers: