$ and \Z work pretty much the same in normal mode, both match the end of search string or before a string-ending newline. the difference between them lies in the multiline mode when you issue an 'm' modifier.

\z means the real end of string even after the string-ending newline.

If you use an 's' modifier, then things become more different but that's mainly coz of the '.' which changes its behaviors, not the three end-of-string anchors..

check the following snippets:

perl -e 'print "match\n" if "foo\n" =~ /.+$/'  # ok #
perl -e 'print "match\n" if "foo\n" =~ /.+\z/'
perl -e 'print "match\n" if "foo\n" =~ /.+\Z/'  # ok #
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\Z/'     
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\z/'    
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+$/'
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+$/m'  # ok #
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\z/m'
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\Z/m'
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\Z/s'  # ok #
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\z/s'  # ok #
perl -e 'print "match\n" if "foo\n\n\n" =~ /.+$/s'  # ok #
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BTW. When comparing between \z, \Z and $, it's probably better to avoid using .* or .? quanifiers the ways in your examples.

BTW. my previous statement about \Z had some error and I have updated that post.

Regards,
Xicheng