# Solve 7th degree equation using complex number

• Nov 23rd 2013, 12:54 AM
anil86
Solve 7th degree equation using complex number
Solve equations using De Moivre's theorem:

1. x^7 + x^4 + x^3 + 1 = 0

2. x^7 - x^4 + x^3 - 1 = 0

I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.
• Nov 23rd 2013, 01:47 AM
chiro
Re: Solve 7th degree equation using complex number
Hey anil86.

Have you been able to factorize the expression in terms (ax + b)^n? (Hint: Consider (x+iy)^n where the real terms are x^7, x^4, x^3 and 1).
• Nov 23rd 2013, 05:25 PM
Soroban
Re: Solve 7th degree equation using complex number
Hello, anil86!

I'll walk you through the first one . . .

Quote:

Solve equations using De Moivre's theorem:

$(1)\;x^7 + x^4 + x^3 + 1 \:=\: 0$

Factor: . $x^4(x^3+1) + (x^3+1) \:=\:0$

Factor: . $(x^3+1)(x^4+1) \:=\:0$

$x^3 + 1 \,=\,0 \quad\Rightarrow\quad x^3 \,=\,-1 \quad\Rightarrow\quad x \,=\,\sqrt[3]{\text{-}1}$

$-1 \:=\:\cos(\pi + 2\pi n) + i\sin(\pi + 2\pi n)$

$(-1)^{\frac{1}{3}} \;=\;\cos\left(\tfrac{\pi}{3} + \tfrac{2\pi}{3}n\right) + i\sin\left(\tfrac{\pi}{3} + \tfrac{2\pi}{3}n\right) \;\text{ for }n = 0,1,2$

$x^4 + 1 \,=\,0 \quad\Rightarrow\quad x^4 \,=\,-1 \quad\Rightarrow\quad x \,=\,\sqrt[4]{\text{-}1}$

$-1 \:=\:\cos(\pi + 2\pi n) + i\sin(\pi + 2\pi n)$

$(-1)^{\frac{1}{4}} \;=\;\cos\left(\tfrac{\pi}{4} + \tfrac{\pi}{2}n\right) + i\sin\left(\tfrac{\pi}{4} + \tfrac{\pi}{2}n\right) \;\text{ for }n = 0,1,2,3$