Complex Number Problem

**CrispyPlanet** · Sep 27th 2013, 10:09 AM

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!

**Idea** · Sep 27th 2013, 10:21 AM

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

**Plato** · Sep 27th 2013, 10:22 AM

Originally Posted by CrispyPlanet

Ok so the problem is as follows:

Find all real number solutions y of:

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

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

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

**HallsofIvy** · Sep 27th 2013, 11:00 AM

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

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

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

**CrispyPlanet** · Sep 27th 2013, 11:19 AM

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

**CrispyPlanet** · Sep 27th 2013, 11:52 AM

Originally Posted by Idea

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

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

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