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How to solve the following equation?

$$\ln(\sin x)-|x|+1=0$$ $x\in\mathbb{C}$

My try:

$$\ln(\sin x)=|x|-1$$

$$f(x):=\ln(\sin x),\\g(x)=|x|-1$$

now:

enter image description here

so Not answer.

is it right?

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    All the graphs show is that there exists no solution over the reals.... Note $x \in \mathbb{C}$. – 2017-02-25
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    @S.C.B. SO how ? – 2017-02-25
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    @Almot1960 I don't know. Hence why it is a comment, not an answer. – 2017-02-25
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    $\pi /2 \pm 3.8610545088374682511449693729817184404319030504907223327437\cdot i$ are solutions – 2017-02-25
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    @Coolwater.How? – 2017-02-25
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    May be this representation be useful $$ e\,\sin(x) =e^{|x|}$$. – 2017-02-25
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    @ why ?$e \sin x=e^{|x| }$ – 2017-02-25
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    $$ \ln (\sin x) =|x|-1 \Rightarrow \sin(x)=e^{|x|-1} \Rightarrow \sin(x)=\frac{e^{|x|}}{e} \Rightarrow e\, \sin(x)=e^{|x|} $$ – 2017-02-25
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    Dose not have [real solutions](http://pasteboard.co/CB64LNRaq.jpg). – 2017-02-25
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    @Amin235 in complex number ? – 2017-02-25

1 Answers 1

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The imaginary part of: $$\ \ \ln(\sin x) -|x|+1\ \ $$ is $$\ \ \arg (\sin (x))\ \ $$ which must be zero, so $\sin(x)$ must be real, non-negative. Your graphs show $x$ is not real, so $x$ must have real part $\pi+2\pi k\ $ for some integer $k$.

Substitute: $x=\pi+2\pi k+b\cdot i$
and rewrite:

$\log (\cosh (b))-\sqrt{(\pi(1+4k)/2)^2+b^2}+1$

which is negative at $b=0$ but goes monotonously to $1-\ln(2)>0$ when $b\to\pm\infty$. So two solutions exist for all $k$. I don't know how to find a closed form for $b$.