How do you:

1. Find all the fifth roots of -1.

2. Find all the third roots of 2+2i

3. Find all the fifth roots of 1 and deduce that:

1 + (2*cos(2*Pi/5)) + (2*cos(4*Pi/5)) = 0.

Help will be very much appreciated.

Thanks

Printable View

- Dec 18th 2007, 03:05 AMsmilerComplex Numbers 1
How do you:

1. Find all the fifth roots of -1.

2. Find all the third roots of 2+2i

3. Find all the fifth roots of 1 and deduce that:

1 + (2*cos(2*Pi/5)) + (2*cos(4*Pi/5)) = 0.

Help will be very much appreciated.

Thanks - Dec 18th 2007, 03:22 AMSimplicity
- Dec 18th 2007, 06:42 AMbobak

other exponents will also give -1. you should be able to figure them out easily.

for number two write the complex number as then solve as you did for number one.

for 3 find the roots using a method simmilar to that used for question 1 and you may identify a geometric series. - Dec 18th 2007, 07:28 AMSoroban
Hello, smiler!

We're expected to know DeMoivre's Theorem: .

Quote:

1. Find all the fifth roots of -1.

. . Then: .

For all five fifth roots: . . . . for

Quote:

2. Find all the third roots of

. . Then: .

For all three cube roots: . . . . for

- Dec 18th 2007, 07:31 AMcolby2152
- Dec 18th 2007, 12:02 PMSimplicity
- Dec 18th 2007, 01:11 PMcolby2152
Wouldn't -1 be a root? (for the 1st question)

- Dec 18th 2007, 09:12 PMbadgerigarQuote:

Wouldn't -1 be a root? (for the 1st question)

No. Sorry, the first part just doesn't work. I think air just misunderstood the fifth roots part of the question (or maybe the question had been edited) - Dec 19th 2007, 06:24 AMcolby2152
Badgerigar... never mind, I found the solution within his at n = 2... That's all I was wondering about. I already know of DeMoivre's imaginary number formula.