# How to determine how many distinct members of set

1. Jan 12, 2008

### sara_87

let
A = {z(belogs to)C | z^6 = −64}
list all members of A
how many distinct members of A are there?

2. Jan 12, 2008

### Rainbow Child

Write $$-64=2^6(\cos(2\,k+1)\,\pi+i\,\sin(2\,k+1)\,\pi)$$, thus

$$z^6=-64\Rightarrow z=2\,\left(\cos\frac{(2\,k+1)\,\pi}{6}+i\,\sin\frac{(2\,k+1)\,\pi}{6}\right)$$

Now it is easy to count the distinct members of A.

3. Jan 12, 2008

### HallsofIvy

Staff Emeritus
The equation zn= a where a is a complex number, always has n distinct solutions in the complex numbers. They are equally spaced around a circle centered on 0 with radius the positive nth root of |a|.

Last edited: Jan 12, 2008
4. Jan 12, 2008

### sara_87

how did you get 2k+1? and how is it easy to count the distinct members?
i'm sorry i know it should be easy but i cant figure out the answer.

5. Jan 12, 2008

### Rainbow Child

The $2\,k+1$ is used in order to write the -1, with trigonometric functions, i.e.

$$\cos(2\,k+1)\,\pi=-1,\quad \sin(2\,k+1)\,\pi=0$$

As HallsofIvy posted there are 6 distinct solutions. If you plug in the equation

$$z=2\,\left(\cos\frac{(2\,k+1)\,\pi}{6}+i\,\sin\frac{(2\,k+1)\,\pi}{6}\right)$$

the values $k=0,1,2,3,4,5,\dots$ then you will see that after the 6th value the solutions, repeat themselves.