In a lecture, the teacher told us that the notion of imaginary number can be introduced using the geometric mean.

We have:

$i=\sqrt{-1}=\sqrt{(-1)\cdot1}$ which is the geometric mean of $-1$ and $1$. We can place $1$ and $-1$ on the real line.

The question is how could you explain that the two geometric mean $i$ and $-i$ are placed on the imaginary line (--EDIT-- without using complex numbers, in order to introduce $i$ ans $-i$).

HINT : In a right triangle, the altitude from the hypotenuse to its 90° vertex splits the hypotenuse into two segments. The geometric mean of these segment lengths is the length of the altitude.

| improve this question | | | | |
  • 1
    $\begingroup$ I'm hoping to understand how this is a useful introduction. The mean of -2 and 8 is 4i. This seems harder to grasp than just introducing it as the square root of a negative number. $\endgroup$ – JTP - Apologise to Monica Nov 21 '14 at 17:59
  • 1
    $\begingroup$ This is certainly not the easiest way to introduce the subject to students, but this is an historical example coming from Jean-Robert Argand. $\endgroup$ – Ortomala Lokni Nov 21 '14 at 18:42
  • 1
    $\begingroup$ This seems to underline the ambiguity of the square root. If we can take the geometric mean of $-2$ and $8$, we should probably also be able to take the geometric mean of $-2$ and $-8$. Is that then supposed to be $-4$ or $4$? How should one choose the sign for $\sqrt{-2\cdot8}$? I would feel tempted to remove this ambiguity by using the square of the geometric mean instead of the geometric mean itself, but this is then just the usual introduction of numbers with negative squares. My guess is that an introduction with the geometric mean would be more confusing than helpful. $\endgroup$ – Joonas Ilmavirta Nov 21 '14 at 21:01

Consider complex numbers in polar form $r e^{i \theta}$. Then multiplying by $re^{i \theta}$ stretches the original number by a factor of $r$ and rotates about the origin by an angle $\theta$. For geometric mean, consider a couple of cases:

1) Two real positive numbers $x=r$ and $y$. For both $\theta = 0$, and the geometric mean is $\sqrt{xy}$. No rotation is require to get from one to the other, only real stretching.

2) Real $x>0$, $y<0$. Then $y = |y|e^{i \pi}$, so $\sqrt{xy} = (x\cdot |y|e^{i \pi})^{1/2} = \sqrt{x|y|}e^{i \pi/2} = i\sqrt{x|y|}$

3) Arbitrary complex $x=r_xe^{i \theta_x}$, $y=r_ye^{i \theta_y}$. Then $\sqrt{xy} = \sqrt{r_xr_y}e^{i(\theta_x+\theta_y)/2}$.

In other words, the geometric mean is the usual geometric mean of the lengths but in the direction halfway between them (i.e. at the arithmetic mean of the angles). Rotating halfway between positives and negatives gives you a purely imaginary answer.

| improve this answer | | | | |
  • $\begingroup$ Your explanation is right. Sorry, if my question was not enough precise but you use complex numbers. How to explain the result of the geometric mean of 1 and -1 to someone that doesn't know complex numbers (in order to introduce i and -i). $\endgroup$ – Ortomala Lokni Nov 21 '14 at 18:57
  • $\begingroup$ Explain that the geometric mean is halfway between turning around. Halfway between right and left is either up or down. We take the convention of up and call this direction i. But I don't know if that's the best way to introduce imaginary numbers... $\endgroup$ – Aeryk Nov 21 '14 at 19:52
  • 2
    $\begingroup$ I love this statement: "Rotating halfway between positives and negatives gives you a purely imaginary answer." $\endgroup$ – rbp Nov 22 '14 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.