All the Fourth Roots of -16

[TiPster]

I'm working on my trig homework and am hopelessly confused!

The problem is to find all four 4th roots of -16.

a⁴=-16 somehow needs to end up being 2cis(45°+k90°) where k=0,1,2,3

I just don't know how to get from point A to point B (figuratively speaking).

 

[DrPhil]

I'm working on my trig homework and am hopelessly confused!

The problem is to find all four 4th roots of -16.

a⁴=-16 somehow needs to end up being 2cis(45°+k90°) where k=0,1,2,3

I just don't know how to get from point A to point B (figuratively speaking).

It isn't a bad idea to think of it as getting from point $\displaystyle A=a^4 = 16 \mathrm{cis} (180°)$ to point $\displaystyle B$ which is the principal 4th root. The procedure (formula) for the nth root is to take the nth root of the amplitude, and to divide the angle by n. More formally, you are thus dividing the natural log of the number by n.

Once you have the principal nth root, all the others are equally spaced around the circle.

Give it a try??

 

[HallsofIvy]

Here is a very basic way to do it: The square roots of 16 are, of course, 4 and -4, and the root of -1 is i: the square roots of -16 are 4i and -4i. The square roots of 4 are 2 and -2 so it only remains to determine the square root of i and -i.

If a+ bi is a square root of i then $\displaystyle (a+ bi)^2= a^2- b^2+ (2ab)i= i$ so that $\displaystyle a^2- b^2= 0$ and 2ab= 1. (a- b)(a+ b)= 0 so either a= b or a= -b. If a= b, $\displaystyle 2ab=2a^2= 1$ and $\displaystyle a= \pm\frac{\sqrt{2}}{2}$ so that the square roots of i are $\displaystyle \frac{2}{2}(1+ i)$ and $\displaystyle -\frac{\sqrt{2}}{2}(1- i)$. Similarly, the two square roots of -i are $\displaystyle \frac{\sqrt{2}}{2}(1- i)$ and $\displaystyle -\frac{\sqrt{2}}{2}$.

Multiplying by 2, the fourth roots of -16 are $\displaystyle \sqrt{2}(1+ i)$, $\displaystyle -\sqrt{2}(1+ i)$, $\displaystyle \sqrt{2}(1- i)$, and $\displaystyle -\sqrt{2}(1- i)$.

 

[Deleted member 4993]

I'm working on my trig homework and am hopelessly confused!

The problem is to find all four 4th roots of -16.

a⁴=-16 somehow needs to end up being 2cis(45°+k90°) where k=0,1,2,3

I just don't know how to get from point A to point B (figuratively speaking).

I would use D'Moiver's therem:

-16 = 16*e^i*(2n+1)*π

(-16)^1/4 = (2⁴*e^i*(2*n+1)*π )^1/4 = 2*[cos{(2n+1)*π/4} + i * sin{(2n+1)*π/4}] for n = 0, 1, 2 & 3