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Question Number 215017 by Hery03 last updated on 25/Dec/24
Re^� soudre dans C l′e^� quation : sin(z) = 2.
$${R}\acute {{e}soudre}\:{dans}\:\mathbb{C}\:{l}'\acute {{e}quation}\:: \\ $$$${sin}\left({z}\right)\:=\:\mathrm{2}. \\ $$
Answered by MrGaster last updated on 25/Dec/24
z=−i ln((√(1−2^2 ))+2i_(simplify) )+2kπi k ∈ Z z=−iln((√(−3))+2i)+2kπi z=−iln(i(√3)+2i)+2kπi z=−iln(i((√3)+2))+2kπi z=−i(ln(i)+ln(.(√3)+2))+2kπi z=−i(((iπ)/2)+ln((√3)+2))+2kπi z=(π/2)−i ln((√3)+2)+2kπi z=(π/2)+(2kπ−ln((√3)+2))i Considering that sin(z) is a periodic function also need to add its conjugate solution: z=(π/2)+(2kπ+ln((√3)+2))i so the complete set of solutions is: z=(π/2)+(2kπ±ln((√3)+2))i k ∈ Z
$${z}=−{i}\:\mathrm{ln}\left(\underset{\mathrm{simplify}} {\underbrace{\sqrt{\mathrm{1}−\mathrm{2}^{\mathrm{2}} }+\mathrm{2}{i}}}\right)+\mathrm{2}{k}\pi{i}\:{k}\:\in\:\mathbb{Z} \\ $$$${z}=−{i}\mathrm{ln}\left(\sqrt{−\mathrm{3}}+\mathrm{2}{i}\right)+\mathrm{2}{k}\pi{i} \\ $$$${z}=−{i}\mathrm{ln}\left({i}\sqrt{\mathrm{3}}+\mathrm{2}{i}\right)+\mathrm{2}{k}\pi{i} \\ $$$${z}=−{i}\mathrm{ln}\left({i}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right)+\mathrm{2}{k}\pi{i} \\ $$$${z}=−{i}\left(\mathrm{ln}\left({i}\right)+\mathrm{ln}\left(.\sqrt{\mathrm{3}}+\mathrm{2}\right)\right)+\mathrm{2}{k}\pi{i} \\ $$$${z}=−{i}\left(\frac{{i}\pi}{\mathrm{2}}+\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right)+\mathrm{2}{k}\pi{i} \\ $$$${z}=\frac{\pi}{\mathrm{2}}−{i}\:\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)+\mathrm{2}{k}\pi{i} \\ $$$${z}=\frac{\pi}{\mathrm{2}}+\left(\mathrm{2}{k}\pi−\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right){i} \\ $$$${C}\mathrm{onsidering}\:\mathrm{that}\:\mathrm{sin}\left({z}\right)\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{periodic}\:\mathrm{function}\:\mathrm{also}\:\mathrm{need}\:\mathrm{to} \\ $$$$\mathrm{add}\:\mathrm{its}\:\mathrm{conjugate}\:\mathrm{solution}: \\ $$$${z}=\frac{\pi}{\mathrm{2}}+\left(\mathrm{2}{k}\pi+\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right){i} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{complete}\:\mathrm{set}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{is}: \\ $$$${z}=\frac{\pi}{\mathrm{2}}+\left(\mathrm{2}{k}\pi\pm\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right){i}\:{k}\:\in\:\mathbb{Z} \\ $$
Commented by MathematicalUser2357 last updated on 26/Dec/24
z=(π/2)+(2kπ+ln((√3)+2))i whereas k∈Z
$${z}=\frac{\pi}{\mathrm{2}}+\left(\mathrm{2}{k}\pi+\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right){i}\:\mathrm{whereas}\:{k}\in\mathbb{Z} \\ $$
Commented by Frix last updated on 26/Dec/24
But where′s the solution path?
$$\mathrm{But}\:\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{path}? \\ $$

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