Find all solutions of the equation sin(z) = 100 Solution No. sin?¹(w) does not m

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No. sin?¹(w) does not mean 1/sin(w). It means the angle whose sin is w. Secondly, z = -iln(sqrt(200ie^iz + 1) cannot be a solution because z appears on both sides. Solutions are of the form z=x+iy where x & y are real values. To solve sin(z)=100 write z as x+iy so that sin(x+iy)=100. (this is what you would do for any RHS, real or complex) Now expand the sin with the normal formula sin(z) = sin(x+iy) = sin(x)cos(iy) + cos(x)sin(iy) = 100 Now use the identities cos(ix)=cosh(x) and sin(iy)=isinh(y). ( these can be derived from the definitions of sin and cos in terms of exp ). Equation becomes sin(x)cosh(y) + icos(x)sinh(y) = 100 … (i) Equating imaginary parts : cos(x)sinh(y)=0 so cos(x)=0 or sinh(y)=0 If sinh(y)=0 then y=0 and (i) becomes sin(x)=100 which is impossible. (x is real) Hence cos(x)=0 which implies that x=np+p/2 for integer n. Subbing in (i) gives (-1)ncosh(y)=100 Since cosh(y)>0 always, n must be even and in this case cosh(y)=100 To solve cosh(y)=100 use calculator to find cosh?¹(100)=±5.2983. ( inspect graph of cosh to see there are two solutions ) Bringing all this together gives complex solutions of sin(z)=100 as z = (4m+1)p/2 ± 5.2983i for m any integer. You can check this with Wolfram http://www.wolframalpha.com/input/?i=sin… Note sin(z)=100 has no real solutions because -1=sin(z)=1 if z is real. Hope that's useful. 1 year ago

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