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The Sin(z) Mapping


Date: 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/   
    
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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