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

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- Dec 18th 2007, 02:05 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, 02:22 AMSimplicity
- Dec 18th 2007, 05:42 AMbobak
$\displaystyle e^{ix} = \cos x +i \sin x$

$\displaystyle e^{i\pi} = \cos \pi +i \sin \pi$

$\displaystyle e^{i\pi} = -1$

$\displaystyle (e^{i\pi})^{\frac{1}{5}} = (-1)^{\frac{1}{5}}$

$\displaystyle e^{i \frac{\pi}{5}} = (-1)^{\frac{1}{5}}$

$\displaystyle \cos \frac{\pi}{5} +i \sin \frac{\pi}{5} = (-1)^{\frac{1}{5}}$

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

for number two write the complex number as $\displaystyle Re^{i \theta}$ 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, 06:28 AMSoroban
Hello, smiler!

We're expected to know DeMoivre's Theorem: .$\displaystyle (\cos\theta + i\sin\theta)^n \;=\;\cos n\theta + i\sin n\theta$

Quote:

1. Find all the fifth roots of -1.

. . Then: .$\displaystyle \left(\cos\pi + i\sin\pi\right)^{\frac{1}{5}} \;=\;\cos\frac{\pi}{5} + i\sin \frac{\pi}{5} $

For all five fifth roots: .$\displaystyle \cos\left(\frac{\pi}{5} + \frac{2n\pi}{5}\right) + i\sin\left(\frac{\pi}{5} + \frac{2n\pi}{5}\right)$ . . . for $\displaystyle n \:=\:0,1,2,3,4$

Quote:

2. Find all the third roots of $\displaystyle 2+2i$

. . Then: .$\displaystyle \left[2^{\frac{3}{2}}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\right]^{\frac{1}{3}} \;=\;2^{\frac{1}{2}}\left[\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right]$

For all three cube roots: .$\displaystyle \sqrt{2}\left[\cos\left(\frac{\pi}{12} +\frac{2n\pi}{3}\right) + i\sin\left(\frac{\pi}{12} + \frac{2n\pi}{3}\right)\right]$ . . . for $\displaystyle n \:=\:0,1,2$

- Dec 18th 2007, 06:31 AMcolby2152
- Dec 18th 2007, 11:02 AMSimplicity
- Dec 18th 2007, 12:11 PMcolby2152
Wouldn't -1 be a root? (for the 1st question)

- Dec 18th 2007, 08:12 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, 05:24 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.