Square root of an imaginary number

Click For Summary

Homework Help Overview

The discussion revolves around the mathematical concept of finding the square root of an imaginary number, specifically 3i, and how to represent it on the complex plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to graph the square root of 3i on the imaginary number axis and question the correct representation of this value. There are discussions about the modulus and angle of complex numbers, as well as the geometric properties of square roots in the complex plane.

Discussion Status

Some participants have provided insights into the geometric interpretation of square roots of complex numbers, referencing polar and exponential forms. There is an ongoing exploration of the correct angles and moduli associated with the square roots of 3i, with no consensus reached yet.

Contextual Notes

Participants are navigating the complexities of representing square roots of imaginary numbers, with some confusion about the placement on the imaginary axis and the implications of using polar coordinates.

Ry122
Messages
563
Reaction score
2
Where on the imaginary number axis do i graph sqrt(3i)? At sqrt3?
 
Physics news on Phys.org
Hi, Ry!
Think in terms of the modulus and the angle that complex numer makes with the real axis.

When you multiply two complex numbers, the resultant number's modulus is the product of the factors' moduli, and its angle the SUM of the the factors' angles to the real axis.
 
Ry122 said:
Where on the imaginary number axis do i graph sqrt(3i)? At sqrt3?

(btw, if you type alt-v, it prints √)

No - that would be (√3)i.

You want (√3)(√i) … though that's not on the imaginary axis.

So your radius is correct, but your modulus (angle) isn't. :smile:
 
Last edited:
Since you say "on the imaginary axis", I assume you mean at [/itex]i\sqrt{3}[/itex], rather than "at \sqrt{3}". No, neither of those is correct since neither of those is the squareroot of 3i: \sqrt{3}^2= 3 and (i\sqrt{3})^2= -3 not 3i.

The square roots (there are, of course, two of them) of 3i is not on the imaginary axis. Square roots, in the complex plane, have a nice geometric property. Are you familiar with de'Moivre's formula? If you write a complex number in polar form, as r (cos(\theta)+ isin(\thet)) or, in exponential form, r e^{i\theta}, then the nth power is r^n(cos(n\theta)+ i sin(n\theta))/. That also holds for fractional powers: the nth root is just that with "n" replaced by "1/n".

In particular, the square root of r(cos(\theta)+ i sin(\theta)) is \sqrt{r}(cos(\theta/2)+ i sin(\theta)/2.

3i lies on the positive imaginary axis, at right angles to the positive real axis, at distance 3 from 0: r= 3, \theta= \pi/2. One of its square roots has r= \sqrt{3} and \theta= \pi/4. Since increasing theta by 2\pi just takes us back to the same point, we can also let \theta= \pi/2+ 2\pi= 5\pi/2 and get 5\pi/4 for the other square root of 3i.

That's the geometric property I mentioned: the two square roots of 3i lie on the line at \pi/4 radians or 45 degrees to the positive real axis, at distance \sqrt{3} from 0, 1 in the first quadrant and the other in the third quadrant.

It is even more interesting for higher roots. You might want to look at
[urlhttp://en.wikipedia.org/wiki/Root_of_unity[/URL]
 
Last edited by a moderator:

Similar threads

Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K