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### The Sin(z) Mapping

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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

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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