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I would like to plot this function:

$$x - {(\cos(x) + i\sin(x))^{ix}} = 0$$

I remember about $\cos(x) + i\sin(x) = e^{ix}$, so this can be written as $$x - {(e^{ix})^{ix}} = 0$$

or maybe better $$x - {e^{ixix}} = 0$$ I know it's not that hard, but my math background is very rusty. I tried with wolfram, but I'm not sure why it's not plotting it. Maybe for the $i$?

Is there another tool I can use?

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    $\begingroup$ Well, what is $i^2$? $\endgroup$
    – Feyre
    Jul 6, 2015 at 16:10
  • $\begingroup$ Try separating $x$ into real and imaginary parts as $x=a+bi$ and treating the real and imaginary parts as coordinate axes. $\endgroup$
    – abiessu
    Jul 6, 2015 at 16:11
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    $\begingroup$ $(e^{ix})^{ix}=e^{-x^2}$. Can you plot $f(x)=x-e^{-x^2}$? Then find the value of $x$ for which $f(x)=0$. $\endgroup$
    – Mark Viola
    Jul 6, 2015 at 16:13
  • $\begingroup$ Also, Mathematica plots this fine, use capital I for i, but {hint}, you don't need that. $\endgroup$
    – Feyre
    Jul 6, 2015 at 16:13
  • $\begingroup$ @Taylor, I changed your edit because $e^{(ix)^2}$ masked an issue the OP has. $\endgroup$
    – user228113
    Jul 6, 2015 at 16:15

2 Answers 2

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We have $f(x)=x-(\cos x+i\sin x)^{ix}=x-e^{-x^2}$. Plot this function and find that $f(x)=0$ when $x\approx. 0.65291862487151$.

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$$x - {(\cos(x) + i\sin(x))^{ix}} = 0\Longleftrightarrow$$ $$x - {(e^{ix})^{ix}} = 0\Longleftrightarrow$$ $$x - e^{ixix} = 0\Longleftrightarrow$$ $$x - e^{(ix)^2} = 0\Longleftrightarrow$$ $$x - e^{i^2x^2} = 0\Longleftrightarrow$$ $$x - e^{-x^2} = 0\Longleftrightarrow$$

(because $x$ is real we can write)

$$e^{x^2}x= 1\Longleftrightarrow$$ $$e^{x^2}= \frac{1}{x}\Longleftrightarrow$$ $$x^2=\frac{\log_{10}(\frac{1}{x})}{\log_{10}(e)}\Longleftrightarrow$$ $$x^2=\frac{\log_{10}(\frac{1}{x})}{\frac{1}{\ln(10)}}\Longleftrightarrow$$ $$x^2=\ln\left(\frac{1}{x}\right)\Longleftrightarrow$$ $$x^2=-\ln(x)\Longleftrightarrow$$ $$x=e^{-\frac{W(2)}{2}}=\sqrt{\frac{W(2)}{2}}$$

$W(x)$ is the Lambert $W$-function

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