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Complex Functions

  • Thread starter m_s_a
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  • #1
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Homework Statement


Find the limit :

http://www.a7bk-a-up.com/pic/uDs93333.bmp
http://www.a7bk-a-up.com/pic/zAP94166.bmp


The Attempt at a Solution


http://www.a7bk-a-up.com/pic/g0Q93947.bmp

Homework Statement








result


The Attempt at a Solution

 

Answers and Replies

  • #2
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Last Qiustion::biggrin:
http://www.a7bk-a-up.com/pic/Nww95795.bmp
 
  • #3
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please wait:zzz:
 
  • #4
Dick
Science Advisor
Homework Helper
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Homework Statement


Find the limit :

http://www.a7bk-a-up.com/pic/uDs93333.bmp
http://www.a7bk-a-up.com/pic/zAP94166.bmp


The Attempt at a Solution


http://www.a7bk-a-up.com/pic/g0Q93947.bmp

Homework Statement








result


The Attempt at a Solution

cos(pi/2)+i*sin(pi/2)=i, not i^(1/2). ???
 
  • #5
Dick
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Last Qiustion::biggrin:
http://www.a7bk-a-up.com/pic/Nww95795.bmp
But what? Isn't the limit still 1?
 
  • #6
91
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cos(pi/2)+i*sin(pi/2)=i, not i^(1/2). ???
By imposing
I said let w=i===>w^2=-1
 
  • #7
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But what? Isn't the limit still 1?
Yes,
Find this limit
 
  • #8
Dick
Science Advisor
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By imposing
I said let w=i===>w^2=-1
I don't understand that at all.
 
  • #9
Dick
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Yes,
Find this limit
Substitute zbar for z in the power series. What's wrong with that?
 
  • #10
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  • #11
Dick
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It doesn't have to be analytical. You've shown using the series (or l'Hopital) that if z_n is a series of complex numbers approaching 0, then sin(z_n)/z_n->1. z_n* is also a series of complex numbers approaching 0. The series expansion holds for ANY z.
 
  • #12
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Like this
http://www.a7bk-a-up.com/pic/ydi48834.bmp
 
  • #13
91
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Infact:
http://www.a7bk-a-up.com/pic/LpB47687.jpg
delta=???
 
  • #14
Dick
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Sure. |(sin(z*)/z*|=|sin(z)/z|. Because (sin(z)/z)*=(sin(z*)/z*) and |z|=|z*|. So you don't need analyticity, correct?
 
  • #15
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Sure. |(sin(z*)/z*|=|sin(z)/z|. Because (sin(z)/z)*=(sin(z*)/z*) and |z|=|z*|. So you don't need analyticity, correct?
Dear: Dick
correct 100%.


Thank you for answering me
Thank you very much:blushing:
In Arabic:
:biggrin:شكرًا جزيلاً
 
  • #16
Dick
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Sure. Sorry, I'm not good at the script. afwan.
 
  • #17
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Sure. Sorry, I'm not good at the script. afwan.
O. My Dod
afwan:surprised
very very Excellent
Rather than to learn English
You have learned Arabic
 
  • #18
Dick
Science Advisor
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shukran, afwan, is about as far as I go. Oh, and salam alekum. That's it. I don't even remember how to count, even though this is a math site. So you might want to keep learning english. :)
 
Last edited:
  • #19
show that :
 

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