Mapping of Functions (Complex Analysis)

[bleedblue1234]

Homework Statement

Show that the function w = e^z maps the shaded rectangle in Fig X one-to-one onto the semi-annulus in Fig y.

Fig x is the rectangle -1<x<1 ; 0<y<(x+pi(i))

Fig y is the semi-annulus such that y>0 and -e<r<-1/e

Homework Equations

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The Attempt at a Solution

I'm not quite sure how to show the function is one-to-one. Any tips would be much appreciated.

 

[Dick]

If e^z1=e^z2, then e^z1/e^z2=1=e^(z1-z2). 0<y<(x+pi(i)) looks a little odd if i is the imaginary unit. You don't write inequalities with complex numbers. What is 'i'?