# De Moivre's Theorem and quadratics

• May 3rd 2010, 12:12 PM
Jgirl689
a) Find all solutions of the equation. (Write your answers in the form a + bi. Enter your answers as a comma-separated list.)
http://www.webassign.net/latexImages...a439b3a7cf.gif

Answers I got for this is -2 + 2i and -2-2i..I assume you just use the quadratic formula and solve..is that right?

b) Find the indicated power using De Moivre's Theorem. http://www.webassign.net/latexImages...427f415f3e.gif

Don't know if I did it right, but I got 1024i for this problem..can someone verify?
• May 3rd 2010, 12:21 PM
Anonymous1
Quote:

Originally Posted by Jgirl689
a) Find all solutions of the equation. (Write your answers in the form a + bi. Enter your answers as a comma-separated list.)
http://www.webassign.net/latexImages...a439b3a7cf.gif

Answers I got for this is -2 + 2i and -2-2i..I assume you just use the quadratic formula and solve..is that right?

b) Find the indicated power using De Moivre's Theorem. http://www.webassign.net/latexImages...427f415f3e.gif

Don't know if I did it right, but I got 1024i for this problem..can someone verify?

Code:

>> syms x; >> solve('x^2 + 2*x + 5 = 0',x)   ans =    - 2*i - 1   2*i - 1
Code:

>> (1+ sqrt(-1))^20 ans =       -1024
• May 3rd 2010, 12:22 PM
rubic
[quote=Jgirl689;506170]a) Find all solutions of the equation. (Write your answers in the form a + bi. Enter your answers as a comma-separated list.)
http://www.webassign.net/latexImages...a439b3a7cf.gif

Answers I got for this is -2 + 2i and -2-2i..I assume you just use the quadratic formula and solve..is that right?

yes, it is

b) Find the indicated power using De Moivre's Theorem. http://www.webassign.net/latexImages...427f415f3e.gif

Don't know if I did it right, but I got 1024i for this problem..can someone verify?

check again!!!!
• May 3rd 2010, 01:19 PM
TheCoffeeMachine
Quote:

Originally Posted by rubic
b) Find the indicated power using De Moivre's Theorem. http://www.webassign.net/latexImages...427f415f3e.gif Don't know if I did it right, but I got 1024i for this problem..

Not right, but close.

$\left(1+i\right)^{20} = \left\{\sqrt{2}\left[\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{ \pi}{4}\right)\right]\right\}^{20}$ $= \left(\sqrt{2}\right)^{20}\left\{\cos\left(\dfrac{ 20\pi}{4}\right)+i\sin\left(\dfrac{20\pi}{4}\right )\right\}$ $= 2^{10}\bigg\{\left(5\pi\right)+i\sin\left(5\pi\rig ht)\bigg\}$ $= 2^{10}\bigg\{\cos\left(\pi\right)+i\sin\left(\pi\r ight)\bigg\} = 2^{10}\left(-1+i(0)\right) = -2^{10}.$
• Jun 21st 2010, 08:05 PM
gauravchauhan
Please help me to find out,A problem on De Moivres theorem
First i apologize if i have posted at wrong place....

The problem is here:

Solve x^8+x^5+x^3+1 by using De Moivres theorem

An example in my book might help(didnt helped me at at all)

x^4-x^3+x^2-x+1

Sol:

The equation can be written as

x^4-x^3+x^2-x+1 = (x^5+1)/(x+1)
Hence the required roots of x^5+1 = 0 are same as those of (x^4-x^3+x^2-x+1)(x^5+1)

The equation (x^5+1) gives x^5 = -1 or x = (-1)^1/5

(......and so on)

Now what my trouble is how i can find equation like (x^5+1) for my problem stated at start of post?

I am newbie and poor in maths so little more explanation will help lot
• Jun 21st 2010, 10:45 PM
sa-ri-ga-ma
Solve x^8+x^5+x^3+1 by using De Moivre`s theorem.

The given problem can be factorized as

x^5(x^3 + 1) + 1(x^3 + 1) = 0

(x^3+1)(x^5+1) = 0

Now solve.
• Jun 22nd 2010, 10:02 PM
gauravchauhan
thanks
sa-ri-ga-ma

I did it....