.

.

Printable View

- Jan 15th 2008, 01:54 PMyellow4321quick complex number question
.

. - Jan 15th 2008, 02:06 PMPlato
There are three possible values of z.

- Jan 15th 2008, 02:23 PMThePerfectHacker
- Jan 15th 2008, 02:39 PMyellow4321
ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?

- Jan 15th 2008, 02:51 PMThePerfectHacker
**Theorem:**The polynomial has roots over . Furthermore, if then are these solutions.

**Proof:**We use a fact that the polynomial can have*at most*zeros. If we let then it is evident that and so . In fact if then . Thus, . Now we show that all are distinct. Suppose that then and so that means because so . Thus, the set is a complete set of zeros for .

**Theorem:**If is a complex number then has exactly zeros and those are .

**Proof:**First the polynomial can have at most zeros so if we can show that these are all zeros (and that they are distinct) it will complete the proof. First is the principle -root defined at , and so . Now if it means . Which means are all zeros. Finally to show there are all distinct it is easy. Suppose where then since it means , as in the previous proof. And so this means that all these zeros are distinct. - Jan 15th 2008, 03:27 PMPlato
- Jan 15th 2008, 03:28 PMyellow4321
ok i sort of understand, and now know the fact that you have n-1 solutions if n is the power so solve for k=0,1,2..n-1 but could you show me how to get to the 5pi/6 and -pi/2 why arent i just added pi to pi/6 to get 7pi/6 or something.

arrrr i see so if i worked in degrees i could just do 360/3 which is 120 then add and minus this 30, then il get the radian form.

you guys are my heros