<div dir="ltr"><div class="gmail_default" style="font-family:verdana,sans-serif;font-size:small">To Sinan: 🙏❤️<br></div><div class="gmail_default" style="font-family:verdana,sans-serif;font-size:small">cx den vaz geçtim... </div><div class="gmail_default" style="font-family:verdana,sans-serif;font-size:small">reel den konuşalım mı? </div><div class="gmail_default" style="font-family:verdana,sans-serif;font-size:small"><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Nov 12, 2024 at 10:49 AM Ali Sinan Sertöz <<a href="mailto:sertoz@listweb.bilkent.edu.tr">sertoz@listweb.bilkent.edu.tr</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><u></u>
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<p>Kızım sana yazıyorum, Yılmaz sen oku! :)</p>
<p>Let z=sqrt( i/(2 n Pi) ). Then -1/z^2=2 n Pi i. (Here sqrt(-1)=i,
and n is any integer).<br>
</p>
<p>Then exp(-1/z^2)=1 for all integers n.</p>
<p>Clearly as n goes to infinity, z goes to zero. <br>
</p>
<p>Also as z goes to zero along the real line exp(-1/z^2) goes to
zero.</p>
<p>So we have the two path test to show that exp(-1/z^2) is not
continuous at the origin. <br>
</p>
<p>Conclusion exp(-1/z^2) is not analytic at the origin.</p>
<p>Now fast forward: This function has an essential singularity at
the origin. In every punctured neighborhood of the origin it takes
every value, with at most one exception, infinitely many times.
This is Big Picard theorem. In our case the exception value is
zero. <br>
</p>
<p>Dersimiz bitti. Haftaya quiz yapacağım! :)<br>
</p>
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Ali Sinan Sertöz <br>
Bilkent Üniversitesi, Matematik Bölümü, 06800 Ankara <br>
Ofis Tel: (312) - 290 1490 <br>
Bölüm Sekreterliği: (312) - 266 4377 <br>
Fax: (312) - 290 1797 <br>
e-posta: <a href="mailto:sertoz@bilkent.edu.tr" target="_blank"> sertoz@bilkent.edu.tr</a> <br>
Anasayfa: <a href="http://sertoz.bilkent.edu.tr/" target="_blank">sertoz.bilkent.edu.tr</a><br>
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