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

Question

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?

Answer 1

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 $x\approx. 0.65291862487151$.

Answer 2

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

(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