The Sin(z) MappingDate: 10/13/98 at 15:33:10 From: Penny Hull Subject: Mapping of complex numbers under the transformation z goes to sinz I have tried to answer this using the definition: sin(z) = 1/2i *(e^iz - e^-iz) Hope you can help. Many thanks. Date: 10/14/98 at 00:46:15 From: Doctor Pete Subject: Re: Mapping of complex numbers under the transformation z goes to sinz Hi, The line y = 1 has the form z = x+I in rectangular coordinates (where "I" is the imaginary unit, Sqrt[-1]). Now, we have (by the regular addition formula): Sin[x+I] = Sin[x]Cos[I] + Cos[x]Sin[I] The idea here is to express the righthand side of the above equation in a form where the real and imaginary parts are separated (why?). Note that: Cos[I] = (Exp[I*I] + Exp[-I*I])/2 = (Exp[-1] + Exp[1])/2 This being a real number, call this constant p. Similarly, Sin[I] = (Exp[I*I] - Exp[-I*I])/(2I) = (Exp[-1] - Exp[1])/(2I) which is a purely imaginary number, call this constant qI, where q is real. It follows that: Sin[x+I] = p*Sin[x] + qI*Cos[x] and when we consider x as a parameter that ranges over the reals, it follows that the transformation f : z -> Sin[z] takes the line y = 1, which is parameterized by: {x(t) = t, y(t) = 1} to: {f(x(t)) = p*Sin[t], f(y(t)) = q*Cos[t]} This is more simply expressed as the pair: x = p*Sin[t] y = q*Cos[t] obviously an ellipse. As a relatively easy exercise, what is the ratio of its width to its height (i.e., what is the ratio p/q)? Answering this question tells you why the answer is an ellipse and not a circle. As for the second question, give it a shot. Guess - are they going to be ellipses as well? (They're not - why?) Here's a hint: Cosh[y] = (Exp[y] + Exp[-y])/2 Sinh[y] = (Exp[y] - Exp[-y])/2 - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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