Square root of an imaginary number

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:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top