MHB Complex Number Problems in Applied Maths

AI Thread Summary
The discussion focuses on solving complex number problems in applied mathematics, specifically proving identities involving cosines and using De Moivre's theorem for polynomial equations. Participants are encouraged to share their attempts to solve the problems, which include equations like x^7 + x^4 + x^3 + 1 = 0 and x^7 - x^4 + x^3 - 1 = 0. Factorization techniques are highlighted as effective methods for tackling these equations. One user successfully solved the problems by factorizing and applying De Moivre's theorem. The thread emphasizes collaborative problem-solving in the context of complex numbers.
anil86
Messages
10
Reaction score
0
This is a thread for complex number problems in applied mathematics.

1. Prove that: 1 + cos x + cos 2x + ...cos (n - 1)x
= {1 - cos x + cos (n - 1)x - cos nx} / 2 (1 - cos x)

= 1/2 + [{sin (n - 1/2)x}/2sin (x/2)]2. If a = cos x + i sin x, b = cos y + i sin y, c = cos z + i sin z, prove that

{(b + c) (c + a) (a + b)}/abc = 8 cos (x - y)/2 cos (y - z)/2 cos (z - x)/2
View attachment 1621View attachment 1621
 

Attachments

  • Image0319.jpg
    Image0319.jpg
    80.3 KB · Views: 146
Mathematics news on Phys.org
I have moved this topic here to our Pre-Calculus sub-forum as it is a better fit than Number Theory.

Can you show what you have tried so our helpers know where you are stuck and what mistake(s) you may be making?
 
Solve equations using De Moivre's theorem:

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

2. x^7 - x^4 + x^3 - 1 = 0I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.
 
anil86 said:
Solve equations using De Moivre's theorem:

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

2. x^7 - x^4 + x^3 - 1 = 0I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.
Those polynomials factorise, e.g. $x^7 + x^4 + x^3 + 1 = x^4(x^3 +1) + x^3+1 = \ldots$.
 
Opalg said:
Those polynomials factorise, e.g. $x^7 + x^4 + x^3 + 1 = x^4(x^3 +1) + x^3+1 = \ldots$.

Hi Opalg,

I solved it by first factorizing & then using De-Moivre theorem as suggested by you. Thank you.

Anil
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top