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Question Number 6908 by Tawakalitu. last updated on 02/Aug/16

sin(Z) = 100    find z

$${sin}\left({Z}\right)\:=\:\mathrm{100} \\ $$$$ \\ $$$${find}\:{z}\: \\ $$

Commented by Rasheed Soomro last updated on 02/Aug/16

Always  −1≤sin(z)≤1  ∀z∈R  ,so  sin(z) = 100 is not possible.  In other words no real solution.

$${Always} \\ $$$$−\mathrm{1}\leqslant{sin}\left({z}\right)\leqslant\mathrm{1}\:\:\forall{z}\in\mathbb{R}\:\:,{so}\:\:{sin}\left({z}\right)\:=\:\mathrm{100}\:{is}\:{not}\:{possible}. \\ $$$${In}\:{other}\:{words}\:{no}\:{real}\:{solution}. \\ $$

Answered by sandy_suhendra last updated on 02/Aug/16

Z is a complex number = x+iy  sin (Z) = sin (x+iy)         100  = sin x.cosh y + i cos x.sinh y  100 + 0.i  = sin x.cosh y + i cos x.sinh y  So ⇒(1)    cos x.sinh y = 0  we get, cos x = 0 ⇒ x = (π/2)+k.2π  and sin x = ±1  or sinh y = 0 ⇒y=0, this is not solution because (x+iy) becomes real  and ⇒(2)   sin x.cosh y = 100, substitute sin x =±1                                        cosh y = ± 100 ⇒ y = ± arc cosh 100    Substitute x and y to Z = x + iy                                                      = ((π/2)+k.2π) ± i arc cosh 100

$${Z}\:{is}\:{a}\:{complex}\:{number}\:=\:{x}+{iy} \\ $$$${sin}\:\left({Z}\right)\:=\:{sin}\:\left({x}+{iy}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{100}\:\:=\:{sin}\:{x}.{cosh}\:{y}\:+\:{i}\:{cos}\:{x}.{sinh}\:{y} \\ $$$$\mathrm{100}\:+\:\mathrm{0}.{i}\:\:=\:{sin}\:{x}.{cosh}\:{y}\:+\:{i}\:{cos}\:{x}.{sinh}\:{y} \\ $$$${So}\:\Rightarrow\left(\mathrm{1}\right)\:\:\:\:{cos}\:{x}.{sinh}\:{y}\:=\:\mathrm{0} \\ $$$${we}\:{get},\:{cos}\:{x}\:=\:\mathrm{0}\:\Rightarrow\:{x}\:=\:\frac{\pi}{\mathrm{2}}+{k}.\mathrm{2}\pi\:\:{and}\:{sin}\:{x}\:=\:\pm\mathrm{1} \\ $$$${or}\:{sinh}\:{y}\:=\:\mathrm{0}\:\Rightarrow{y}=\mathrm{0},\:{this}\:{is}\:{not}\:{solution}\:{because}\:\left({x}+{iy}\right)\:{becomes}\:{real} \\ $$$${and}\:\Rightarrow\left(\mathrm{2}\right)\:\:\:{sin}\:{x}.{cosh}\:{y}\:=\:\mathrm{100},\:{substitute}\:{sin}\:{x}\:=\pm\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cosh}\:{y}\:=\:\pm\:\mathrm{100}\:\Rightarrow\:{y}\:=\:\pm\:{arc}\:{cosh}\:\mathrm{100} \\ $$$$ \\ $$$${Substitute}\:{x}\:{and}\:{y}\:{to}\:{Z}\:=\:{x}\:+\:{iy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{\pi}{\mathrm{2}}+{k}.\mathrm{2}\pi\right)\:\pm\:{i}\:{arc}\:{cosh}\:\mathrm{100} \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 02/Aug/16

Wow ... Thanks for your help

$${Wow}\:...\:{Thanks}\:{for}\:{your}\:{help} \\ $$

Commented by Yozzii last updated on 03/Aug/16

If y∈R, coshy≥1>0⇒∄y∈R for coshy=−100.  In fact, sin((π/2)+2kπ)=sin(π/2)=1 for all k∈Z  due to periodicity of sinusoidal function.

$${If}\:{y}\in\mathbb{R},\:{coshy}\geqslant\mathrm{1}>\mathrm{0}\Rightarrow\nexists{y}\in\mathbb{R}\:{for}\:{coshy}=−\mathrm{100}. \\ $$$${In}\:{fact},\:{sin}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right)={sin}\frac{\pi}{\mathrm{2}}=\mathrm{1}\:{for}\:{all}\:{k}\in\mathbb{Z} \\ $$$${due}\:{to}\:{periodicity}\:{of}\:{sinusoidal}\:{function}. \\ $$$$ \\ $$

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