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Rocatiwb

Rocatiwb

Answered question

2022-05-28

Solving z = i ln ( c i + ( 1 + c ) ( 1 c ) ) for c
Given,
z = i ln ( c i + ( 1 + c ) ( 1 c ) )
I would like to solve for c. First multiply through by i,
i z = ln ( c i + ( 1 + c ) ( 1 c ) ) .
Then take the exponential of both sides,
e i z = c i + ( 1 + c ) ( 1 c )
Not sure where to go from here, can I say
( e i z ) = ( c i + ( 1 + c ) ( 1 c ) ) sin z = c ?

Answer & Explanation

ketHideniw7

ketHideniw7

Beginner2022-05-29Added 6 answers

e i z + c i = 1 c 2 e 2 i z + 2 i e i z c c 2 = 1 c 2
e 2 i z + 2 i e i z c = 1 c = 1 e 2 i z 2 i e i z

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