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I tried unsuccessfully to solve the following complex expression and get the module and the argument.

${e}^{ix}+{e}^{2ix}$

I converted the whole expression to trigonometric function cos and sin but it got more complex than it looks at the origin.

SAM.Am
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3 Answers3

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$$ \begin{align} e^{ix}+e^{2ix} &=e^{\frac32ix}\left(e^{\frac12ix}+e^{-\frac12ix}\right)\\ &=e^{i\color{#090}{\frac32x}}\color{#C00}{2\cos\left(\tfrac12x\right)} \end{align} $$ So $$ \left|\,e^{ix}+e^{2ix}\,\right|=\left|\,\color{#C00}{2\cos\left(\tfrac12x\right)}\,\right| $$ and $$ \arg\left(e^{ix}+e^{2ix}\right)\equiv\color{#090}{\tfrac32x}+\pi\left[\,\cos\left(\tfrac12x\right)\lt0\,\right]\pmod{2\pi} $$ where $[\cdots]$ are Iverson Brackets.

robjohn
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  • But $$\cos\dfrac x2$$ can be $$<0?$$ – lab bhattacharjee Sep 20 '19 at 08:37
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    Very nice solution ! –  Sep 20 '19 at 08:37
  • Elegant Solution . Really Beautiful and Beyond my Imagin{ation}{ary} Before I Discovered it. Thanks . – SAM.Am Sep 20 '19 at 08:44
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    @labbhattacharjee: good point! I have added a correction for that – robjohn Sep 20 '19 at 08:48
  • @robjohn, Thanks for the Iverson bracket which is new to me. Actually I was thinking in the same line as well.Regarding argument I think, we need https://en.wikipedia.org/wiki/Atan2#Definition_and_computation . For example, $x=\dfrac{17\pi}3$ – lab bhattacharjee Sep 20 '19 at 08:55
  • @labbhattacharjee: arg is always mod $2\pi$, so I have noted that, too. This way, one can handle $-\pi/4$ and $7\pi/4$ as one sees fit. – robjohn Sep 20 '19 at 09:15
  • What if I use Maclaurin Series. I got this so far:

    $e^{ix} +e^{2ix}=2+ \sum_{k=1}^{\infty} \frac{x^{2k}(1-2^{2k})}{(2k)!} +{i}\sum_{k=0}^{\infty} \frac{x^{2k+1}(1-2^{2k+1})}{(2k+1)!} $

     – SAM.Am Sep 20 '19 at 12:08
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$$1+e^{ix}=1+\cos x+i\sin x=2\cos^2\frac x2+2i\cos\frac x2\sin\frac x2=2\cos\frac x2e^{ix/2}$$

so that

$$e^{ix}+e^{2ix}=2\cos\frac x2 e^{3ix/2}.$$

Notice that the first locus is a shifted circle, while the second is the good old Pascal's limaçon, obtained by tripling the argument.

enter image description here

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Let $$Z=e^{ix}+e^{2ix}=e^{ix}(1+e^{ix}) \Rightarrow |Z|=|1+\cos x +i \sin x|= \sqrt{1+\cos x)^2+\sin ^2x}$$ $$\Rightarrow |Z|=\sqrt{2+2\cos x}=2 \cos(x/2).$$

$$Arg(Z) =Arg[(\cos x+\cos 2x)+i(\sin x+ \sin 2x)]= \tan^{-1} \frac{\sin x+\sin 2x}{\cos x + \cos 2x}$$ $$ \Rightarrow Arg(Z)= \tan^{-1} \tan (3x/2)=3x/2.$$

Z Ahmed
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