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Question Number 204417 by Frix last updated on 16/Feb/24
Solve for z∈C e^z =ln z
$$\mathrm{Solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{e}^{{z}} =\mathrm{ln}\:{z} \\ $$
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
$${say}\:{z}={re}^{\theta{i}} ={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${with}\:{r},\:\theta\:\in\mathbb{R}\:{and}\:{r}\geqslant\mathrm{0}. \\ $$$${e}^{{r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)} =\mathrm{ln}\:{r}+\theta{i} \\ $$$${e}^{{r}\mathrm{cos}\:\theta} {e}^{\left({r}\:\mathrm{sin}\:\theta\right){i}} =\mathrm{ln}\:{r}+\theta{i} \\ $$$${e}^{{r}\mathrm{cos}\:\theta} \left[\mathrm{cos}\:\left({r}\:\mathrm{sin}\:\theta\right)+{i}\:\mathrm{sin}\:\left({r}\:\mathrm{sin}\:\theta\right)\right]=\mathrm{ln}\:{r}+\theta{i} \\ $$$$\begin{cases}{{e}^{{r}\:\mathrm{cos}\:\theta} \mathrm{cos}\:\left({r}\:\mathrm{sin}\:\theta\right)=\mathrm{ln}\:{r}}\\{{e}^{{r}\:\mathrm{cos}\:\theta} \mathrm{sin}\:\left({r}\:\mathrm{sin}\:\theta\right)=\theta}\end{cases} \\ $$$${one}\:{of}\:{infinite}\:{solutions}\:{is}: \\ $$$$\theta\approx\pm\mathrm{1}.\mathrm{33724},\:{r}\approx\mathrm{1}.\mathrm{37456} \\ $$
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 (x)^(1/n) =c e^x =a+bi versus ln x =a+bi
$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{are}\:\mathrm{only}\:\mathrm{2}\:\mathrm{solutions}. \\ $$$$\mathrm{The}\:\mathrm{problems}\:\mathrm{are}\:\mathrm{very}\:\mathrm{similar}: \\ $$$${x}^{{n}} ={c}\:\mathrm{versus}\:\sqrt[{{n}}]{{x}}={c} \\ $$$$\mathrm{e}^{{x}} ={a}+{b}\mathrm{i}\:\mathrm{versus}\:\mathrm{ln}\:{x}\:={a}+{b}\mathrm{i} \\ $$
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 ⇒ z=e^(−W(−1)) ∨z=e^(−W_(−1) (−1)) (z≈.318132±1.33724i) 2 solutions z=e^z ⇒ z=−W_n (−1); n∈Z infinitely many solutions e^z =ln z ⇒ z≈.318132±1.33724i 2 solutions
$$\mathrm{Wolframalpha}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{following}\:\mathrm{solutions}: \\ $$$${z}=\mathrm{ln}\:{z}\:\Rightarrow\:{z}=\mathrm{e}^{−{W}\left(−\mathrm{1}\right)} \vee{z}=\mathrm{e}^{−{W}_{−\mathrm{1}} \left(−\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\left({z}\approx.\mathrm{318132}\pm\mathrm{1}.\mathrm{33724i}\right) \\ $$$$\mathrm{2}\:\mathrm{solutions} \\ $$$$ \\ $$$${z}=\mathrm{e}^{{z}} \:\Rightarrow\:{z}=−{W}_{{n}} \left(−\mathrm{1}\right);\:{n}\in\mathbb{Z} \\ $$$$\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solutions} \\ $$$$ \\ $$$$\mathrm{e}^{{z}} =\mathrm{ln}\:{z}\:\Rightarrow\:{z}\approx.\mathrm{318132}\pm\mathrm{1}.\mathrm{33724i} \\ $$$$\mathrm{2}\:\mathrm{solutions} \\ $$
Commented by mr W last updated on 17/Feb/24
each of the intersection points from the green curves and the red curves represents a solution. e.g. z=2.4206e^(±6.9223i)
$${each}\:{of}\:{the}\:{intersection}\:{points}\:{from} \\ $$$${the}\:{green}\:{curves}\:{and}\:{the}\:{red}\:{curves} \\ $$$${represents}\:{a}\:{solution}. \\ $$$${e}.{g}.\:\:{z}=\mathrm{2}.\mathrm{4206}{e}^{\pm\mathrm{6}.\mathrm{9223}{i}} \\ $$
Answered by Frix last updated on 17/Feb/24
e^a =b ⇔ a=ln b Let a=b=z e^z =z ⇔ z=ln z ⇔ e^z =ln z ⇒ We can solve e^z =z and test our solutions e^z =z z=a+bi e^a (cos b +i sin b)=a+bi sin b =be^(−a) cos b =ae^(−a) ⇒ tan b =(b/a) ⇔ a=(b/(tan b)) sin b =be^(−(b/(tan b))) ⇔ b−e^(b/(tan b)) sin b =0 b_1 ≈±1.33723570143 ⇒ a_1 ≈.318131505205 z_1 ≈.318131505205±1.33723570143i z_1 ≈1.37455701074e^(±1.33723570143i) Test e^z_1 =z_1 true ln z_1 =z_1 true b_2 ≈±7.58863117847 ⇒ a_2 ≈2.06227772960 z_2 ≈2.06227772960±7.58863117847i z_2 ≈7.86386117609e^(±1.30544587129i) Test e^z_2 =z_2 true ln z_2 ≈2.06227772960±1.30544587129i ≠ z_2 false
$$\mathrm{e}^{{a}} ={b}\:\Leftrightarrow\:{a}=\mathrm{ln}\:{b} \\ $$$$\mathrm{Let}\:{a}={b}={z} \\ $$$$\mathrm{e}^{{z}} ={z}\:\Leftrightarrow\:{z}=\mathrm{ln}\:{z}\:\Leftrightarrow\:\mathrm{e}^{{z}} =\mathrm{ln}\:{z} \\ $$$$\Rightarrow\:\mathrm{We}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\mathrm{e}^{{z}} ={z}\:\mathrm{and}\:\mathrm{test}\:\mathrm{our}\:\mathrm{solutions} \\ $$$$ \\ $$$$\mathrm{e}^{{z}} ={z} \\ $$$${z}={a}+{b}\mathrm{i} \\ $$$$\mathrm{e}^{{a}} \left(\mathrm{cos}\:{b}\:+\mathrm{i}\:\mathrm{sin}\:{b}\right)={a}+{b}\mathrm{i} \\ $$$$\mathrm{sin}\:{b}\:={b}\mathrm{e}^{−{a}} \\ $$$$\mathrm{cos}\:{b}\:={a}\mathrm{e}^{−{a}} \\ $$$$\Rightarrow\:\mathrm{tan}\:{b}\:=\frac{{b}}{{a}}\:\Leftrightarrow\:{a}=\frac{{b}}{\mathrm{tan}\:{b}} \\ $$$$\mathrm{sin}\:{b}\:={b}\mathrm{e}^{−\frac{{b}}{\mathrm{tan}\:{b}}} \:\Leftrightarrow\:{b}−\mathrm{e}^{\frac{{b}}{\mathrm{tan}\:{b}}} \mathrm{sin}\:{b}\:=\mathrm{0} \\ $$$$ \\ $$$${b}_{\mathrm{1}} \approx\pm\mathrm{1}.\mathrm{33723570143}\:\Rightarrow\:{a}_{\mathrm{1}} \approx.\mathrm{318131505205} \\ $$$${z}_{\mathrm{1}} \approx.\mathrm{318131505205}\pm\mathrm{1}.\mathrm{33723570143i} \\ $$$${z}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{37455701074e}^{\pm\mathrm{1}.\mathrm{33723570143i}} \\ $$$$\mathrm{Test} \\ $$$$\mathrm{e}^{{z}_{\mathrm{1}} } ={z}_{\mathrm{1}} \:\mathrm{true} \\ $$$$\mathrm{ln}\:{z}_{\mathrm{1}} ={z}_{\mathrm{1}} \:\mathrm{true} \\ $$$$ \\ $$$${b}_{\mathrm{2}} \approx\pm\mathrm{7}.\mathrm{58863117847}\:\Rightarrow\:{a}_{\mathrm{2}} \approx\mathrm{2}.\mathrm{06227772960} \\ $$$${z}_{\mathrm{2}} \approx\mathrm{2}.\mathrm{06227772960}\pm\mathrm{7}.\mathrm{58863117847i} \\ $$$${z}_{\mathrm{2}} \approx\mathrm{7}.\mathrm{86386117609e}^{\pm\mathrm{1}.\mathrm{30544587129i}} \\ $$$$\mathrm{Test} \\ $$$$\mathrm{e}^{{z}_{\mathrm{2}} } ={z}_{\mathrm{2}} \:\mathrm{true} \\ $$$$\mathrm{ln}\:{z}_{\mathrm{2}} \:\approx\mathrm{2}.\mathrm{06227772960}\pm\mathrm{1}.\mathrm{30544587129i}\:\neq\:{z}_{\mathrm{2}} \:\mathrm{false} \\ $$

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