sinz=z   ,  z=x+iy

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beckden | High School Teacher | (Level 1) Educator

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sin w=z
sin w=((e^{iw}-e^{iw})/(2i))
e^{iw}-e^{-iw}=2iz   multiply everything by e^{iw}
e^{2iw}-1=2ize^{iw}
e^{2iw}-2ize^{iw}-1=0
y=e^{iw}
y²-2izy-1=0
y=((2iz±√((-2iz)²-4(-1)))/2)=((-iz±√(-4z²+4))/2)
e^{iw}=-iz±√(1-z²)   so our answer is
iw=ln(-iz±√(1-z²))
w=(1/i)ln(-iz±√(1-z²))+2πn

I do not see an easy way to solve this equation for z.

z=0 is an obvious solution.

Using newton's method on the original function

f(z) = sin(z)-z

f '(z) = cos(z) - 1

Pick a x_0 and iterate the following formula

x_(n+1) = x_n - f(x_n)/f '(x_n)

I get z = 7.4976763 - 2.7686783 i as one posible answer from about 10 iterations starting with (5+5i)

z = 13.899960 - 3.3522099 i  is another answer.

I do not see a pattern, but perhaps you can...

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