Square root of an imaginary number

Click For Summary
SUMMARY

The square roots of the imaginary number 3i are located at a distance of √3 from the origin in the complex plane, specifically at angles of π/4 and 5π/4 radians relative to the positive real axis. Utilizing de Moivre's formula, the square root can be expressed in polar form as √3(cos(θ/2) + i sin(θ/2)). The discussion clarifies that the square roots of 3i do not lie on the imaginary axis, but rather at specific angles in the complex plane.

PREREQUISITES
  • Understanding of complex numbers and their polar representation
  • Familiarity with de Moivre's theorem
  • Knowledge of modulus and argument of complex numbers
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study de Moivre's theorem in detail
  • Learn about polar coordinates and their application in complex analysis
  • Explore the geometric interpretation of complex roots
  • Investigate the concept of roots of unity and their significance
USEFUL FOR

Mathematicians, physics students, and anyone interested in complex analysis or the geometric properties of complex numbers.

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
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K