Solve a complex equation in R

Question

Solve

$x=1+w^x + w^{2x}$

Where $w$ ,$w^2$ are cube roots of the unit and $x$ is pure real number My attempt :-

let $w = \cos(120) + i \sin(120)$

Then

$x = 1+\cos(120x) + \cos(240x) + [\sin(120x) + \sin(240x) ]i$

$x = 1+\cos(120x) + \cos(240x) $

And

$\sin(120x) + \sin(240x) = 0$

I can solve the second equation i fo not know how can i solve the first one ?

If you mean $120$ to be degrees you should include the degree sign, which you can get with ^\circ. Otherwise you can use $\frac 23\pi$. Also putting a backslash before common functions gives the correct font and spacing, so \sin x gives $\sin x$ instead of sin x which gives $sin x$ — Ross Millikan, Jun 09 '20 at 23:43

J. W. Tanner · Answer 1 · 2020-06-09T23:37:58.157

1

$x=1+w^x+w^{2x}\implies x(w^x-1)=w^{3x}-1=0\implies x=0 $ or $w^x=1$.

But $x=0$ is not a solution, so $x=1+w^x+w^{2x}=1+1+1=3$.

edited Jun 09 '20 at 23:37

answered Jun 09 '20 at 23:21

J. W. Tanner

60,406

score 0 · Answer 2 · answered Jun 10 '20 at 00:00

0

Since $w^2=1/w$, we have $$x=1+w^x+\frac{1}{w^x} \iff xw^x=1+w^x+w^{2x}=x$$ Now $x=0$ isn't a solution so $w^x=1 \implies x=3k$, where $k$ is a non-zero integer, but this means $3k=1+1+1 \implies k=1, x=3$

answered Jun 10 '20 at 00:00

Anas A. Ibrahim

1,884
7
15

Solve a complex equation in R

2 Answers2