# Complex Numbers 1

• Dec 18th 2007, 02:05 AM
smiler
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
• Dec 18th 2007, 02:22 AM
Simplicity
Quote:

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

$\sqrt{-1} = i$

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

Can you do the rest?
• Dec 18th 2007, 05:42 AM
bobak
$e^{ix} = \cos x +i \sin x$
$e^{i\pi} = \cos \pi +i \sin \pi$
$e^{i\pi} = -1$
$(e^{i\pi})^{\frac{1}{5}} = (-1)^{\frac{1}{5}}$
$e^{i \frac{\pi}{5}} = (-1)^{\frac{1}{5}}$
$\cos \frac{\pi}{5} +i \sin \frac{\pi}{5} = (-1)^{\frac{1}{5}}$

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

for number two write the complex number as $Re^{i \theta}$ 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 AM
Soroban
Hello, smiler!

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

Quote:

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

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

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

Quote:

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

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

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

• Dec 18th 2007, 06:31 AM
colby2152
Quote:

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

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

Can you do the rest?

I believe Air meant to write this:

$^5\sqrt {-1} = i^5 = i$
• Dec 18th 2007, 11:02 AM
Simplicity
Quote:

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

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

Oh yes... I was considering as $5 \times \sqrt{-1} = 5i$.
• Dec 18th 2007, 12:11 PM
colby2152
Wouldn't -1 be a root? (for the 1st question)
• Dec 18th 2007, 08:12 PM