Solve for z∈C e^z =ln z


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Question Number 204417 by Frix last updated on 16/Feb/24

$Solve for z \in C$ $e^{z} = \ln z$
	Answered by mr W last updated on 17/Feb/24
	$say z=re^(θi) =r(cos θ+i sin θ) with r, θ ∈R and r≥0. e^(r(cos θ+i sin θ)) =ln r+θi e^(rcos θ) e^((r sin θ)i) =ln r+θi e^(rcos θ) [cos (r sin θ)+i sin (r sin θ)]=ln r+θi { ((e^(r cos θ) cos (r sin θ)=ln r)),((e^(r cos θ) sin (r sin θ)=θ)) :} one of infinite solutions is: θ≈±1.33724, r≈1.37456$
	$s a y z = {r e}^{θ i} = r (\cos θ + i \sin θ)$ $w i t h r, θ \in R a n d r ⩾ 0 .$ $e^{r (\cos θ + i \sin θ)} = \ln r + θ i$ $e^{r \cos θ} e^{(r \sin θ) i} = \ln r + θ i$ $e^{r \cos θ} [\cos (r \sin θ) + i \sin (r \sin θ)] = \ln r + θ i$ ${\begin{cases} e^{r \cos θ} \cos (r \sin θ) = \ln r \\ e^{r \cos θ} \sin (r \sin θ) = θ \end{cases}$ $o n e o f i n f i n i t e s o l u t i o n s i s :$ $θ \approx \pm 1.33724, r \approx 1.37456$
	Commented by Frix last updated on 17/Feb/24

	$I think there are only 2 solutions .$ $The problems are very similar :$ $x^{n} = c versus \sqrt[n]{x} = c$ $e^{x} = a + b i versus \ln x = a + b i$
	Commented by mr W last updated on 17/Feb/24

	Commented by Frix last updated on 18/Feb/24

	$Wolframalpha gives the following solutions :$ $z = \ln z \Rightarrow z = e^{- W (- 1)} \lor z = e^{- W_{- 1} (- 1)}$ $(z \approx . 318132 \pm 1 . 33724 i)$ $2 solutions$ $z = e^{z} \Rightarrow z = - W_{n} (- 1); n \in Z$ $infinitely many solutions$ $e^{z} = \ln z \Rightarrow z \approx . 318132 \pm 1 . 33724 i$ $2 solutions$
	Commented by mr W last updated on 17/Feb/24

	$e a c h o f t h e i n t e r s e c t i o n p o i n t s f r o m$ $t h e g r e e n c u r v e s a n d t h e r e d c u r v e s$ $r e p r e s e n t s a s o l u t i o n .$ $e . g . z = 2.4206 e^{\pm 6.9223 i}$
	Answered by Frix last updated on 17/Feb/24

	$e^{a} = b \Leftrightarrow a = \ln b$ $Let a = b = z$ $e^{z} = z \Leftrightarrow z = \ln z \Leftrightarrow e^{z} = \ln z$ $\Rightarrow We can solve$ $e^{z} = z and test our solutions$ $e^{z} = z$ $z = a + b i$ $e^{a} (\cos b + i \sin b) = a + b i$ $\sin b = b e^{- a}$ $\cos b = a e^{- a}$ $\Rightarrow \tan b = \frac{b}{a} \Leftrightarrow a = \frac{b}{\tan b}$ $\sin b = b e^{- \frac{b}{\tan b}} \Leftrightarrow b - e^{\frac{b}{\tan b}} \sin b = 0$ $b_{1} \approx \pm 1.33723570143 \Rightarrow a_{1} \approx . 318131505205$ $z_{1} \approx . 318131505205 \pm 1 . 33723570143 i$ $z_{1} \approx 1 . {37455701074 e}^{\pm 1 . 33723570143 i}$ $Test$ $e^{z_{1}} = z_{1} true$ $\ln z_{1} = z_{1} true$ $b_{2} \approx \pm 7.58863117847 \Rightarrow a_{2} \approx 2.06227772960$ $z_{2} \approx 2.06227772960 \pm 7 . 58863117847 i$ $z_{2} \approx 7 . {86386117609 e}^{\pm 1 . 30544587129 i}$ $Test$ $e^{z_{2}} = z_{2} true$ $\ln z_{2} \approx 2.06227772960 \pm 1 . 30544587129 i \neq z_{2} false$
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