Homework Help: Complex Functions

m_s_a · Jun 12, 2008

1. The problem statement, all variables and given/known data
Find the limit :

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

3. The attempt at a solution
http://www.a7bk-a-up.com/pic/g0Q93947.bmp
1. The problem statement, all variables and given/known data

result

3. The attempt at a solution

m_s_a · Jun 12, 2008

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

m_s_a · Jun 15, 2008

please wait:zzz:

Dick · Jun 16, 2008

m_s_a said: ↑

1. The problem statement, all variables and given/known data
Find the limit :

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

3. The attempt at a solution
http://www.a7bk-a-up.com/pic/g0Q93947.bmp
1. The problem statement, all variables and given/known data

result

3. The attempt at a solution

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

Dick · Jun 16, 2008

m_s_a said: ↑

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

But what? Isn't the limit still 1?

m_s_a · Jun 17, 2008

Dick said: ↑

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 · Jun 17, 2008

Dick said: ↑

But what? Isn't the limit still 1?

Yes,
Find this limit

Dick · Jun 17, 2008

m_s_a said: ↑

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

I don't understand that at all.

Dick · Jun 17, 2008

m_s_a said: ↑

Yes,
Find this limit

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

m_s_a · Jun 17, 2008

Dick said: ↑

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 · Jun 17, 2008

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 · Jun 17, 2008

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

m_s_a · Jun 17, 2008

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

Dick · Jun 17, 2008

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 · Jun 17, 2008

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
In Arabic:
شكرًا جزيلاً

Dick · Jun 17, 2008

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

m_s_a · Jun 17, 2008

Dick said: ↑

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

Dick · Jun 17, 2008

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 · Sep 24, 2008

show that :