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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.

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  • $\begingroup$ Is the module the absolute value ? Your approach is good. Why do you expect an easy formula for module and argument ? $\endgroup$ – Peter Sep 20 at 8:24
  • $\begingroup$ for modulus, see this question $\endgroup$ – J. W. Tanner Sep 20 at 11:32
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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.

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  • $\begingroup$ But $$\cos\dfrac x2$$ can be $$<0?$$ $\endgroup$ – lab bhattacharjee Sep 20 at 8:37
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    $\begingroup$ Very nice solution ! $\endgroup$ – Yves Daoust Sep 20 at 8:37
  • $\begingroup$ Elegant Solution . Really Beautiful and Beyond my Imagin{ation}{ary} Before I Discovered it. Thanks . $\endgroup$ – SAM.Am Sep 20 at 8:44
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    $\begingroup$ @labbhattacharjee: good point! I have added a correction for that $\endgroup$ – robjohn Sep 20 at 8:48
  • $\begingroup$ @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 en.wikipedia.org/wiki/Atan2#Definition_and_computation . For example, $x=\dfrac{17\pi}3$ $\endgroup$ – lab bhattacharjee Sep 20 at 8:55
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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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    $\begingroup$ It's really a pleasure to read a such Gem . Thanks . $\endgroup$ – SAM.Am Sep 20 at 9:06
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Let $$Z=e^{ix}+e^{2ix}=e^{ix}(1+e^{ix}) \Rightarrow |Z|=|1+\cos x +i \sin x|= \sqrt{1+\cos x)^2+\sin ^2x}$$ $$\Rightarrow |Z|=\sqrt{2+2\cos x}=2 \cos(x/2).$$

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

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  • $\begingroup$ Another piece of art . Elegant too. $\endgroup$ – SAM.Am Sep 20 at 8:49

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