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Where on the imaginary number axis do i graph sqrt(3i)? At sqrt3?

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- Thread starter Ry122
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Where on the imaginary number axis do i graph sqrt(3i)? At sqrt3?

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arildno

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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.

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tiny-tim

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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

So your radius is correct, but your modulus (angle) isn't.

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HallsofIvy

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Since you say "on the imaginary axis", I assume you mean at [/itex]i\sqrt{3}[/itex], rather than "at [itex]\sqrt{3}[/itex]". No, neither of those is correct since neither of those is the squareroot of 3i: [itex]\sqrt{3}^2= 3[/itex] and [itex](i\sqrt{3})^2= -3[/itex] 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 [itex]r (cos(\theta)+ isin(\thet))[/itex] or, in exponential form, [itex]r e^{i\theta}[/itex], then the n^{th} power is [itex]r^n(cos(n\theta)+ i sin(n\theta))/[/itex]. 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 [itex]r(cos(\theta)+ i sin(\theta))[/itex] is [itex]\sqrt{r}(cos(\theta/2)+ i sin(\theta)/2[/itex].

3i lies on the positive imaginary axis, at right angles to the positive real axis, at distance 3 from 0: r= 3, [itex]\theta= \pi/2[/itex]. One of its square roots has [itex]r= \sqrt{3}[/itex] and [itex]\theta= \pi/4[/itex]. Since increasing [itex]theta[/itex] by [itex]2\pi[/itex] just takes us back to the same point, we can also let [itex]\theta= \pi/2+ 2\pi= 5\pi/2[/itex] and get [itex]5\pi/4[/itex] 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 [itex]\pi/4[/itex] radians or 45 degrees to the positive real axis, at distance [itex]\sqrt{3}[/itex] 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]

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 [itex]r (cos(\theta)+ isin(\thet))[/itex] or, in exponential form, [itex]r e^{i\theta}[/itex], then the n

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

3i lies on the positive imaginary axis, at right angles to the positive real axis, at distance 3 from 0: r= 3, [itex]\theta= \pi/2[/itex]. One of its square roots has [itex]r= \sqrt{3}[/itex] and [itex]\theta= \pi/4[/itex]. Since increasing [itex]theta[/itex] by [itex]2\pi[/itex] just takes us back to the same point, we can also let [itex]\theta= \pi/2+ 2\pi= 5\pi/2[/itex] and get [itex]5\pi/4[/itex] 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 [itex]\pi/4[/itex] radians or 45 degrees to the positive real axis, at distance [itex]\sqrt{3}[/itex] 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]

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