Solve 2 cosh ⁡ z + sinh ⁡ z = i for z

letumsnemesislh

letumsnemesislh

Answered

2022-07-04

Solve 2 cosh z + sinh z = i for z

Answer & Explanation

gozaderaradiox5

gozaderaradiox5

Expert

2022-07-05Added 19 answers

2 cosh ( z ) + sinh ( z ) = i
e z + e z + e z e z 2 = i
2 e z + 2 e z + e z e z = 2 i
3 e z + e z = 2 i
3 e z + ( e z ) 1 = 2 i
Let e z = x
3 x + x 1 = 2 i
3 x 2 + 1 = 2 i x
3 x 2 2 i x + 1 = 0
We solve equations of the form a x 2 + b x + c = 0 with
x 1 , 2 = b ± b 2 4 a c 2 a
By plugging in a=3, b=−2i and c=1:
x 1 , 2 = 2 i ± 4 4 3 1 6
= 2 i ± 4 12 6 = 2 i ± 16 6 = 2 i ± 4 i 6 = i 3 ± 2 i 3
The two roots are therefore
x 1 = i , x 2 = i 3
Since we required e z = x, z = Ln ( x ) (Ln being the complex-valued logarithm). We are lazy and throw both values in Wolfram Alpha and get
z 1 = ln ( i ) = i π 2 , z 2 = ln ( i 3 ) = ln ( 3 ) i π 2
And now we have z 1 and z 2 , which are the solutions to the initial equation. Notice that the solution to that complex logarithm has arbitrary multiples of 2 π i in it

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