Re^� soudre dans C l′e^� quation : sin(z) = 2.


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Question Number 215017 by Hery03 last updated on 25/Dec/24

$R \overset{´}{e s o u d r e} d a n s C l^{'} \overset{´}{e q u a t i o n} :$ $s i n (z) = 2 .$
	Answered by MrGaster last updated on 25/Dec/24

	$z = - i \ln (\underset{simplify}{\underset{⏟}{\sqrt{1 - 2^{2}} + 2 i}}) + 2 k π i k \in Z$ $z = - i \ln (\sqrt{- 3} + 2 i) + 2 k π i$ $z = - i \ln (i \sqrt{3} + 2 i) + 2 k π i$ $z = - i \ln (i (\sqrt{3} + 2)) + 2 k π i$ $z = - i (\ln (i) + \ln (. \sqrt{3} + 2)) + 2 k π i$ $z = - i (\frac{i π}{2} + \ln (\sqrt{3} + 2)) + 2 k π i$ $z = \frac{π}{2} - i \ln (\sqrt{3} + 2) + 2 k π i$ $z = \frac{π}{2} + (2 k π - \ln (\sqrt{3} + 2)) i$ $C onsidering that \sin (z) is a$ $periodic function also need to$ $add its conjugate solution :$ $z = \frac{π}{2} + (2 k π + \ln (\sqrt{3} + 2)) i$ $so the complete set of solutions is :$ $z = \frac{π}{2} + (2 k π \pm \ln (\sqrt{3} + 2)) i k \in Z$
	Commented by MathematicalUser2357 last updated on 26/Dec/24

	$z = \frac{π}{2} + (2 k π + \ln (\sqrt{3} + 2)) i whereas k \in Z$
	Commented by Frix last updated on 26/Dec/24

	$But {where}^{'} s the solution path ?$
	Commented by mr W last updated on 27/Dec/24

	$w r o n g!$ $n o t z = \frac{π}{2} + (2 k π + \ln (\sqrt{3} + 2)) i$ $b u t z = \frac{π}{2} + 2 k π + (\ln (\sqrt{3} + 2)) i$
	Answered by mr W last updated on 27/Dec/24

	$s a y z = a + b i w i t h b \neq 0$ $\sin (a + b i) = \sin a \cos (b i) + \cos a \sin (b i) = 2$ $\sin a \cosh b + i \cos a \sinh b = 2$ $\underset{―}{\cos a \sinh b = 0}$ $\Rightarrow \cos a = 0 \Rightarrow a = 2 k π \pm \frac{π}{2}$ $\underset{―}{\sin a \cosh b = 2}$ $w i t h a = 2 k π + \frac{π}{2} :$ $\Rightarrow \sin a = 1$ $\Rightarrow \cosh b = 2 \Rightarrow b = \cosh^{- 1} 2 = \ln (2 + \sqrt{3})$ $w i t h a = 2 k π - \frac{π}{2} :$ $\Rightarrow \sin a = - 1$ $\Rightarrow \cosh b = - 2 \Rightarrow i m p o s s b l e$ $s u m m a r y :$ $z = 2 k π + \frac{π}{2} + i \ln (2 + \sqrt{3})$
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