Solving a Periodic FunctionDate: 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... Rich 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/