Complex Functions Homework: Find Limit

[m_s_a]

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

 

[m_s_a]

Last Qiustion:
http://www.a7bk-a-up.com/pic/Nww95795.bmp

 

[m_s_a]

please wait:zzz:

 

[Dick]

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). ?

 

[Dick]

Last Qiustion:
http://www.a7bk-a-up.com/pic/Nww95795.bmp

But what? Isn't the limit still 1?

 

[m_s_a]

cos(pi/2)+i*sin(pi/2)=i, not i^(1/2). ?

By imposing
I said let w=i===>w^2=-1

 

[m_s_a]

But what? Isn't the limit still 1?

Yes,
Find this limit

 

[Dick]

By imposing
I said let w=i===>w^2=-1

I don't understand that at all.

 

[Dick]

Yes,
Find this limit

Substitute zbar for z in the power series. What's wrong with that?

 

[m_s_a]

Substitute zbar for z in the power series. What's wrong with that?

Oh No , This can not ; Because sinz*/z* Not Analytical
http://en.wikipedia.org/wiki/Analytic_function

 

[Dick]

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.

 

[m_s_a]

Like this
http://www.a7bk-a-up.com/pic/ydi48834.bmp

 

[m_s_a]

Infact:
http://www.a7bk-a-up.com/pic/LpB47687.jpg
delta=?

 

[Dick]

Sure. |(sin(z*)/z*|=|sin(z)/z|. Because (sin(z)/z)*=(sin(z*)/z*) and |z|=|z*|. So you don't need analyticity, correct?

 

[m_s_a]

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
In Arabic:
شكرًا جزيلاً

 

[Dick]

Sure. Sorry, I'm not good at the script. afwan.

 

[m_s_a]

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

 

[Dick]

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. :)

 

[salemghazou]

show that :