# Thread: Complex Numbers 1

1. ## Complex 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

2. Originally Posted by smiler
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
$\displaystyle \sqrt{-1} = i$

So for 1.) $\displaystyle 5\sqrt {-1} = 5i$

Can you do the rest?

3. $\displaystyle e^{ix} = \cos x +i \sin x$
$\displaystyle e^{i\pi} = \cos \pi +i \sin \pi$
$\displaystyle e^{i\pi} = -1$
$\displaystyle (e^{i\pi})^{\frac{1}{5}} = (-1)^{\frac{1}{5}}$
$\displaystyle e^{i \frac{\pi}{5}} = (-1)^{\frac{1}{5}}$
$\displaystyle \cos \frac{\pi}{5} +i \sin \frac{\pi}{5} = (-1)^{\frac{1}{5}}$

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.

4. Hello, smiler!

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

1. Find all the fifth roots of -1.
Convert $\displaystyle -1 + 0i$ to polar form: .$\displaystyle (1,\,\pi) \quad\Rightarrow\quad \cos\pi + i\sin\pi$

. . 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 $\displaystyle n \:=\:0,1,2,3,4$

2. Find all the third roots of $\displaystyle 2+2i$
$\displaystyle 2 + 2i \quad\Rightarrow\quad \left(2\sqrt{2},\,\frac{\pi}{4}\right) \quad\Rightarrow\quad 2^{\frac{3}{2}}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

. . 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 $\displaystyle n \:=\:0,1,2$

5. Originally Posted by Air
$\displaystyle \sqrt{-1} = i$

So for 1.) $\displaystyle 5\sqrt {-1} = 5i$

Can you do the rest?
I believe Air meant to write this:

$\displaystyle ^5\sqrt {-1} = i^5 = i$

6. Originally Posted by colby2152
I believe Air meant to write this:

$\displaystyle ^5\sqrt {-1} = i^5 = i$
Oh yes... I was considering as $\displaystyle 5 \times \sqrt{-1} = 5i$.

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

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

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)

9. 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.