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e-z-1-3-i-z-




Question Number 48224 by gunawan last updated on 21/Nov/18
e^z =1−(√3)i  z=..
ez=13iz=..
Answered by Smail last updated on 21/Nov/18
z=ln(1−(√3)i)=ln(2((1/2)−i((√3)/2)))  =ln2+ln(e^(−i(π/3)) )=ln2−i(π/3)
z=ln(13i)=ln(2(12i32))=ln2+ln(eiπ3)=ln2iπ3

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