# Complex Functions

## 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

result

## The Attempt at a Solution

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

Dick
Homework Helper

## 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

result

## The Attempt at a Solution

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

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

But what? Isn't the limit still 1?

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

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

But what? Isn't the limit still 1?

Yes,
Find this limit

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

I don't understand that at all.

Dick
Homework Helper
Yes,
Find this limit

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

Dick
Homework Helper
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.

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

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

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

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 very much
In Arabic:
شكرًا جزيلاً

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

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