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 n
th power is r^n(cos(n\theta)+ i sin(n\theta))/. That also holds for fractional powers: the n
th 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
[url
http://en.wikipedia.org/wiki/Root_of_unity[/URL]