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
$\displaystyle e^{ix} = \cos x +i \sin x$
$\displaystyle e^{i\pi} = \cos \pi +i \sin \pi$
$\displaystyle e^{i\pi} = -1$
$\displaystyle (e^{i\pi})^{\frac{1}{5}} = (-1)^{\frac{1}{5}}$
$\displaystyle e^{i \frac{\pi}{5}} = (-1)^{\frac{1}{5}}$
$\displaystyle \cos \frac{\pi}{5} +i \sin \frac{\pi}{5} = (-1)^{\frac{1}{5}}$
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.
Hello, smiler!
We're expected to know DeMoivre's Theorem: .$\displaystyle (\cos\theta + i\sin\theta)^n \;=\;\cos n\theta + i\sin n\theta$
Convert $\displaystyle -1 + 0i$ to polar form: .$\displaystyle (1,\,\pi) \quad\Rightarrow\quad \cos\pi + i\sin\pi$1. Find all the fifth roots of -1.
. . Then: .$\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 $\displaystyle n \:=\:0,1,2,3,4$
$\displaystyle 2 + 2i \quad\Rightarrow\quad \left(2\sqrt{2},\,\frac{\pi}{4}\right) \quad\Rightarrow\quad 2^{\frac{3}{2}}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) $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 $\displaystyle n \:=\:0,1,2$
Yes it would. For a method of finding all 5 of the complex roots, see soroban's excellent post (you may need to google de moivres theorem if you haven't seen this kind of question before).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)