Complex but Simple equation. Im( e^z ) = 1 I've reached rsin(t) = 1 but I am stuck here, where r is radius and t is argument. There may be use of logs here I might be missing.

Jenny Schroeder

Jenny Schroeder

Answered question

2022-11-05

Complex but Simple equation.
Im( e^z ) = 1
I've reached rsin(t) = 1 but I am stuck here, where r is radius and t is argument.
There may be use of logs here I might be missing.

Answer & Explanation

ustalovatfog

ustalovatfog

Beginner2022-11-06Added 11 answers

Assuming z = a + b i with a , b R :
( e a + b i ) = 1
( i e a + b i ) = 1
i ( e a + b i + ( e a + b i ) ) = 1
i e a + b i + i ( e a + b i ) = 1
i e a + b i + i e a cos ( b ) = 1
e a sin ( b ) = 1
Solving the last part we get:
a = ln ( csc ( b ) ) + 2 i π n
b = π arcsin ( 1 e a ) + 2 π n         b = arcsin ( 1 e a ) + 2 π n
With n Z

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