Why are there 3 solutions to (1 + isqrt(3))^(1/3)) ??

Question

THe question I'm working on is

Compute (1+i sqrt(3))^1/3 in the form of rcis(t)

So I know that
z^n=r^n(cis(nt))

So I've worked out r = 2 so r^(1/3) = 2^(1/3)

and arctan (sqrt(3)) = pi/3

Now the answer I would then put is

rcis(t) = 2^(1/3) cis (pi/3) AND 2^(1/3) cis (-pi/3)

(because it's a cubed root you have to include the negative... or at least thats the part I'm not sure on)

Because the solutions I've got have a 3rd solution

2^(1/3) cis(pi)

How is that so?

Answer 1

As you probably know a complex number can be written in the form:

z=r*e^(ia), and e^(ia) = cos a + i sin a

for each integer n:
e^(ia + 2n*pi) = cos(a + 2n*pi) + i sin (+ 2n*pi) = cos a + i sin a

Now, 2*cos(pi/3) = 1,
2*sin(pi/3) = sqrt(3)

So, it is ok to take r=2, t=pi/3

1+i sqrt(3) = 2 * e^(i * pi /3) =
= e^[i * (2pi + pi/3)] = e^[i * (4pi + pi/3)] = ...

[1+i sqrt(3)]^(1/3) = 2^(1/3) * e^[(i * pi /3)/3] =
= 2^(1/3)* e*(i*pi/9)

Other roots are 2^(1/3) * e^[i/3 * (2pi + pi/3)] =
= 2^(1/3) * e^[i/3 * 7pi/3] = 2^(1/3) * e^(i*7pi/9)

2^(1/3) * e^[i/3 * (4pi + pi/3)] = 2^(1/3) * e^[i/3 * 13pi/3]=
= 2^(1/3) * e^(i*13pi/9)

Answer 2

Every nth root has n roots (solutions). Although there are n roots, often these roots exist in the complex plane. For example, 8^1/3 has 3 roots; however, we often only think of the "real roots" (a term I'm sure you've commonly heard). In addition, these roots will always be separated by a constant angle in the complex plane. Furthermore, these roots will lie on a circle of radius |z|^1/n, where z is 8 in the above example. I'm not sure if this is what you're looking for, but if you need more explanation, please let me know.

Answer 3

I'm not following all of your notation, but I'll tell you this.

x^2 = 1 has two solutions
cos(0) + i sin(0) = 1
cos(2Pi/2) + i sin(2Pi/2) = -1

x^3 = 1 has three solutions, and these solutions are

cos(0) + i sin(0)
cos(2Pi/3)+i sin(2Pi/3)
cos(4Pi/3)+i sin(4Pi/3)

Any of these values when cubed gives you one.

x^4 = 1 gives four solutions:
cos(0)+i sin(0)
cos(2Pi/4)+i sin(2Pi/4) = i
cos(4Pi/4)+i sin(4Pi/4) = -1
cos(6Pi/4)+i sin(4Pi/4) = - i

An easier way to think of complex numbers is as a magnitude and an angle. When you multiply them, the angles add together. If a complex number goes 1/3, 2/3, or 3/3 of the way around the circle, cubing it goes around the circle once, twice or three times... Since it is a whole number of times, it becomes a real number.