Module & argument of complex number exp(ix) +exp(2ix) 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. 
 A: $$
\begin{align}
e^{ix}+e^{2ix}
&=e^{\frac32ix}\left(e^{\frac12ix}+e^{-\frac12ix}\right)\\
&=e^{i\color{#090}{\frac32x}}\color{#C00}{2\cos\left(\tfrac12x\right)}
\end{align}
$$
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.
A: $$1+e^{ix}=1+\cos x+i\sin x=2\cos^2\frac x2+2i\cos\frac x2\sin\frac x2=2\cos\frac x2e^{ix/2}$$
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.

A: Let $$Z=e^{ix}+e^{2ix}=e^{ix}(1+e^{ix}) \Rightarrow |Z|=|1+\cos x +i \sin x|= \sqrt{1+\cos x)^2+\sin ^2x}$$ $$\Rightarrow |Z|=\sqrt{2+2\cos x}=2 \cos(x/2).$$
$$Arg(Z) =Arg[(\cos x+\cos 2x)+i(\sin x+ \sin 2x)]= \tan^{-1} \frac{\sin x+\sin 2x}{\cos x + \cos 2x}$$ $$ \Rightarrow Arg(Z)= \tan^{-1} \tan (3x/2)=3x/2.$$
