Complex Functions Homework: Find Limit

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The discussion revolves around finding the limit of a complex function, with participants sharing their attempts and solutions. Key points include the use of trigonometric identities, such as cos(pi/2) and sin(pi/2), and the application of series expansions to evaluate limits. There is a debate on the necessity of analyticity in the context of the limit, with some arguing that it is not required. Participants also emphasize the validity of using series expansions for any complex number approaching zero. The conversation concludes with expressions of gratitude and light-hearted exchanges about language skills.
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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

 
Physics news on Phys.org
Last Qiustion::biggrin:
http://www.a7bk-a-up.com/pic/Nww95795.bmp
 
please wait:zzz:
 
m_s_a said:

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). ?
 
m_s_a said:
Last Qiustion::biggrin:
http://www.a7bk-a-up.com/pic/Nww95795.bmp

But what? Isn't the limit still 1?
 
Dick said:
cos(pi/2)+i*sin(pi/2)=i, not i^(1/2). ?

By imposing
I said let w=i===>w^2=-1
 
Dick said:
But what? Isn't the limit still 1?

Yes,
Find this limit
 
m_s_a said:
By imposing
I said let w=i===>w^2=-1

I don't understand that at all.
 
m_s_a said:
Yes,
Find this limit

Substitute zbar for z in the power series. What's wrong with that?
 
  • #10
  • #11
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
Like this
http://www.a7bk-a-up.com/pic/ydi48834.bmp
 
  • #13
Infact:
http://www.a7bk-a-up.com/pic/LpB47687.jpg
delta=?
 
  • #14
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
Dick said:
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
Sure. Sorry, I'm not good at the script. afwan.
 
  • #17
Dick said:
Sure. Sorry, I'm not good at the script. afwan.

O. My Dod
afwan
very very Excellent
Rather than to learn English
You have learned Arabic
 
  • #18
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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