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Question Number 57719 by Joel578 last updated on 10/Apr/19
Findallcomplexnumberzthatsatisfysinhz=i
Answered by mr W last updated on 10/Apr/19
sinhz=ez−e−z2=i⇒(ez)2−2i(ez)−1=0⇒(ez−i)2=0⇒ez=i=cos(2nπ+π2)+isin(2nπ+π2)=e(2nπ+π2)i⇒z=(2nπ+π2)i
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