Question Number 205147 by Ghisom last updated on 10/Mar/24
$$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$${z}\mathrm{ln}\:{z}\:={z}−\mathrm{2} \\ $$
Answered by pi314 last updated on 10/Mar/24
$${z}={e}^{{y}} \\ $$$$\Leftrightarrow{ye}^{{y}} ={e}^{{y}} −\mathrm{2} \\ $$$$\left({y}−\mathrm{1}\right){e}^{{y}−\mathrm{1}} =−\mathrm{2}{e}^{−\mathrm{1}} \\ $$$${y}−\mathrm{1}={W}\left(−\mathrm{2}{e}^{−\mathrm{1}} \right) \\ $$$${y}=\mathrm{1}+{W}\left(−\mathrm{2}{e}^{−\mathrm{1}} \right) \\ $$$${z}={e}^{\mathrm{1}+{W}\left(−\mathrm{2}{e}^{−\mathrm{1}} \right)} \\ $$$$ \\ $$
Commented by Frix last updated on 10/Mar/24
$$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{approximate}\:\mathrm{value}\:\mathrm{of}\:{z}? \\ $$
Answered by Frix last updated on 10/Mar/24
$$\mathrm{Different}\:\mathrm{method}: \\ $$$${z}\mathrm{ln}\:{z}\:−{z}+\mathrm{2}=\mathrm{0} \\ $$$${z}\left(\mathrm{ln}\:{z}\:−\mathrm{1}\right)+\mathrm{2}=\mathrm{0} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$${r}\mathrm{e}^{\mathrm{i}\theta} \left(\mathrm{ln}\:{r}\:−\mathrm{1}+\mathrm{i}\theta\right)+\mathrm{2}=\mathrm{0} \\ $$$$\begin{cases}{{r}\left(\mathrm{cos}\:\theta\:\mathrm{ln}\:{r}\:−\mathrm{cos}\:\theta\:−\theta\mathrm{sin}\:\theta\right)+\mathrm{2}=\mathrm{0}}\\{\mathrm{i}{r}\left(\mathrm{sin}\:\theta\:\mathrm{ln}\:{r}\:+\theta\mathrm{cos}\:\theta\:−\mathrm{sin}\:\theta\right)=\mathrm{0}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:{r}=\mathrm{e}^{\mathrm{1}−\frac{\theta}{\mathrm{tan}\:\theta}} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}−\mathrm{e}^{\mathrm{1}−\frac{\theta}{\mathrm{tan}\:\theta}} \frac{\theta}{\mathrm{sin}\:\theta}=\mathrm{0} \\ $$$$\theta\approx\pm\mathrm{1}.\mathrm{13267249760} \\ $$$${r}\approx\mathrm{1}.\mathrm{59896034818} \\ $$$${z}\approx.\mathrm{678344933205}\pm\mathrm{1}.\mathrm{44793727304i} \\ $$