Question

Trigonometric form of negative square root 3 +i ?

Answer 1

There are two square roots of `3+i`,

• `root{4}{10} exp(i/2 tan^-1(1/3))`, and
• `- root{4}{10} exp(i/2 tan^-1(1/3))`

Since there is no order relation among complex numbers - there is really no sense to speaking about the "negative" square root.

Explanation:

The complex number `3+i` can be put in the trigonometric form (I am more comfortable calling it the polar form - so that is what I will do from now on) by choosing real numbers `r` and `theta` such that

`rcos theta = 3, qquad r sin theta = 1`

These are easily solved by

`r = sqrt((r costheta)^2+(r sin theta)^2)= sqrt10,qquad theta = tan^-1(1/3)`

So,

`3+i = re^{i theta} = sqrt 10 exp(i tan^-1(1/3))`

However, changing `theta` by any integer multiple of `2pi` does not change the number, so we can write

`3+i = re^{i (theta+2pi n)} = sqrt 10 exp(i (tan^-1(1/3)+2 pi n)), n in ZZ`

The square root of `3+i` is, then

`sqrt(re^{i (theta+2pi n)})=sqrt(r)e^{i(theta/2+pi n)}`
`qquad = root{4}{10} exp(i/2 (tan^-1(1/3)+2 pi n))`
`qquad = (-1)^nroot{4}{10} exp(i/2 tan^-1(1/3))`

where we have used `e^{i pi}=-1`

Thus, there are two square roots of `3+i`,

• `root{4}{10} exp(i/2 tan^-1(1/3))`, and
• `- root{4}{10} exp(i/2 tan^-1(1/3))`

Since there is no order relation among complex numbers - there is really no sense to speaking about the "negative" square root. You can not really say that a complex number is less than zero, hence negative (although I would guess it is the second of the two square roots above that is being meant here)!