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Solving a Periodic Function

Date: 3/31/96 at 10:45:50
From: Anonymous
Subject: complex analysys

sin2z = 5

I tried 2sinzcosz = 2*(1/2i)(e^iz - e^-iz)*(1/2)(e^iz + e^-iz) = 5

after simplification

1/2i(e^i2z - e^-i2z) = 5

I know sin is periodic on 2pi. and e^z is never zero...
where next...?

Please help...

Date: 5/30/96 at 23:30:2
From: Doctor Pete
Subject: Re: complex analysys

First, notice you don't really need to use the sine double-angle 
formula.  Say n = 2z.  Then 
sin(n) = (e^(ni)-e^(-ni))/(2i) = (e^(2zi)-e^(-2zi))/(2i) = 5, 
as you found below.  Then we can continue our solution as follows:

        e^(2zi)-e^(-2zi) = 10i

multiply both sides by e^(2zi):

     => e^(2zi)*(e^(2zi)-e^(-2zi)) = e^(2zi)*10i
     => (e^(2zi))^2-1 = 10i*e^(2zi)

Let w=e^(2zi).  Then:

        w^2-1 = 10i*w
     => w^2-10i*w-1 = 0

which is a quadratic in w.  Using the quadratic equation, we find

        w = (10 +- sqrt(6))i/2 ,

where +- is the plus/minus symbol.  Then remembering the 
substitution made, we have

        z = Log(w)/(2i)
          = (ln|w|+i*arg(w)+2*Pi*i*k)/(2i)
          = (ln((10+-sqrt(6))/2)+i*Pi*(1/2+2k)/(2i),

for any integer k.  So we have 2 infinite families of solutions.

The key here in this solution is recognizing that the exponential 
form of the sine is actually a quadratic; the motivation for this 
comes from solving for arcsinh(x), given sinh(x):

        sinh(x) = (e^x-e^-x)/2 = y
     => e^x-e^-x = 2y
     => (e^x)^2-1 = 2y*e^x
     => e^x = (-2y +- sqrt(4y^2+4))/2
            = -y +- sqrt(y^2+1)
     => x = ln(-y +- sqrt(y^2+1)).

-Doctor Pete,  The Math Forum

Associated Topics:
High School Trigonometry

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