Thread: Complex Number Problem

Ok so the problem is as follows:

Find all real number solutions y of:

e^(iy) + e^(2iy) + e^(3iy)=0

Firstly I tried letting e^(iy) equal some variable X. Then:

X + X^2 + X^3=0 ==> X(1 + X + X^2)=0.

I used the quadratic formula on (1 + X + X^2) to get X=-1/2 ± (sqrt(3)/2)*i

So e^(iy)=-1/2 ± (sqrt(3)/2)*i

Where do I go from here? Am I even on the right track? Any help much appreciated!

$\displaystyle e^{i y}+e^{2i y}+e^{3i y}=e^{2i y}\left(1+e^{i y}+e^{-i y}\right)=e^{2i y}(1+2\cos y)$

Originally Posted by CrispyPlanet

Ok so the problem is as follows:

Find all real number solutions y of:

$\displaystyle e^{iy} + e^{2iy} + e^{3iy}=0$

That is: $\displaystyle \cos(y)+\cos(2y)+\cos(3y)+i(\sin(y)+\sin(2y)+\sin( 3y))=0$

So you need $\displaystyle \cos(y)+\cos(2y)+\cos(3y)= 0~\&~\sin(y)+\sin(2y)+\sin(3y)= 0$

Idea and Plato have used different methods but what you have done is perfectly valid.

Of course, $\displaystyle iy= ln(-1/2+ i\sqrt{3}/2$, in the first case, and $\displaystyle iy= ln(-1/2- i\sqrt{3}/2$. To find that logarithm, write the numbers in "polar" or exponential form: $\displaystyle \sqrt{(-1/2)^2+ (\sqrt{3}/2)^2}= \sqrt{1/4+ 3/4}= 1$ and $\displaystyle arctan(\frac{\sqrt{3}/2}{1/2})= arctan(\sqrt{3})= \frac{\pi}{3}$ so that we have $\displaystyle ln(e^{i\pi/3})= i\pi/3$ so that $\displaystyle iy= i\pi/3$ and then $\displaystyle y= \pi/3$.

The logarithm, like most functions extended to the complex numbers, is not "single valued". $\displaystyle e^{i(x+ 2k\pi)}= e^{ix}$, for any integer k, so that $\displaystyle ln(re^{ix})= ln(re^{i(x+ 2k\pi)}= ln(r)+ (x+ 2\pi) i$. that is, we can add any multiple of $\displaystyle 2\pi$ to the solution above.

Hi HallsofIvy,

When I plug in pi/3 into the polar formula (cosy+cos2y+cos3y+i(siny+sin2y+sin3y)=0), the result is not zero. What am I missing? I've tried working with the cosines and sines separately, making them equal to zero, but I'm not getting anywhere with that either.

CP

Originally Posted by Idea

$\displaystyle e^{i y}+e^{2i y}+e^{3i y}=e^{2i y}\left(1+e^{i y}+e^{-i y}\right)=e^{2i y}(1+2\cos y)$

Hi Idea, many thanks for your response. I worked out your method and got y=2pi/3, which makes the equation equal to zero. So the solution is y=(±2pi/3 +2k*pi).
What I'm not certain about is why my method yielded y=pi/3.

CP