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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?
$$\mathrm{But}\:\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{path}? \\ $$
Commented by mr W last updated on 27/Dec/24
wrong! not z=(π/2)+(2kπ+ln((√3)+2))i but z=(π/2)+2kπ+(ln((√3)+2))i
$${wrong}! \\ $$$${not}\:{z}=\frac{\pi}{\mathrm{2}}+\left(\mathrm{2}{k}\pi+\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right){i}\: \\ $$$${but}\:{z}=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi+\left(\mathrm{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)\right){i}\: \\ $$
Answered by mr W last updated on 27/Dec/24
say z=a+bi with b≠0 sin (a+bi)=sin a cos (bi)+cos a sin (bi)=2 sin a cosh b+i cos a sinh b=2 cos a sinh b=0 ⇒cos a=0 ⇒a=2kπ±(π/2) sin a cosh b=2 with a=2kπ+(π/2): ⇒sin a=1 ⇒cosh b=2 ⇒b=cosh^(−1) 2=ln (2+(√3)) with a=2kπ−(π/2): ⇒sin a=−1 ⇒cosh b=−2 ⇒impossble summary: z=2kπ+(π/2)+i ln (2+(√3))
$${say}\:{z}={a}+{bi}\:{with}\:{b}\neq\mathrm{0} \\ $$$$\mathrm{sin}\:\left({a}+{bi}\right)=\mathrm{sin}\:{a}\:\mathrm{cos}\:\left({bi}\right)+\mathrm{cos}\:{a}\:\mathrm{sin}\:\left({bi}\right)=\mathrm{2} \\ $$$$\mathrm{sin}\:{a}\:\mathrm{cosh}\:{b}+{i}\:\mathrm{cos}\:{a}\:\mathrm{sinh}\:{b}=\mathrm{2} \\ $$$$\underline{\mathrm{cos}\:{a}\:\mathrm{sinh}\:{b}=\mathrm{0}\:} \\ $$$$\Rightarrow\mathrm{cos}\:{a}=\mathrm{0}\:\Rightarrow{a}=\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$$$\underline{\mathrm{sin}\:{a}\:\mathrm{cosh}\:{b}=\mathrm{2}} \\ $$$${with}\:{a}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}: \\ $$$$\Rightarrow\mathrm{sin}\:{a}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{cosh}\:{b}=\mathrm{2}\:\Rightarrow{b}=\mathrm{cosh}^{−\mathrm{1}} \:\mathrm{2}=\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$${with}\:{a}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}}: \\ $$$$\Rightarrow\mathrm{sin}\:{a}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cosh}\:{b}=−\mathrm{2}\:\Rightarrow{impossble} \\ $$$${summary}: \\ $$$${z}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}+{i}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$

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