Solving Complex Analysis Questions: Are My Answers Right?

[mick25]

Homework Statement

I just wrote a test and was wondering if I got these questions right, I already solved them, please see the attached pictures below. Here are the questions; sorry for non-latex form

1) Let gamma be a positively oriented unit circle (|z|=1) in C

solve: i) integral of cos(z+11)/(z^2-23) along gamma
ii) integral of cos(z+11)/(z^2-6z) along gamma

2) Find all possible analytic functions f(z) such that |sin 7z| < |f(z)|

3) Draw a closed curve gamma in C\<0,1> such that its winding numbers at 0 is 1 and at 1 is -2.

4) Draw 4 pairwise non-homotopic closed curves in C\<0>

The Attempt at a Solution

Here are the pictures:

Are these correct? Please let me know if my writing is ineligible. Thanks in advance.

 

[jackmell]

Your writing is illegible to me. Not too hard to do latex. Let's see if I can write the first one:

$$\oint \frac{\cos(z+11)}{z^2+23} dz,\quad z=e^{it}$$

Looks all analytic in there so that's zero right? Think you said that. Now if you want, do a quote on my post to see the Latex code and if you want to post more in here, try and learn how to do it.

 

[mick25]

Your writing is illegible to me. Not too hard to do latex. Let's see if I can write the first one:

$$\oint \frac{\cos(z+11)}{z^2+23} dz,\quad z=e^{it}$$

Looks all analytic in there so that's zero right? Think you said that. Now if you want, do a quote on my post to see the Latex code and if you want to post more in here, try and learn how to do it.

It's actually

$$\oint \frac{\cos(z+11)}{z^2-23} dz$$

and since $$z = ±√23$$ is not in the unit circle, it must be 0 right?

The second integral is

$$\oint \frac{\cos(z+11)}{z^2-6z} dz$$

which can be written as

$$\oint \frac{\cos(z+11)}{z(z-6)} dz$$

then

$$\oint \frac{\frac{\cos(z+11)}{z-6}}{z} dz$$

so evaluating $$\oint \frac{\cos(z+11)}{z-6} dz at z=0$$ gives $$\frac{pi*icos(11)}{3}$$

 

[jackmell]

It's actually

$$\oint \frac{\cos(z+11)}{z^2-23} dz$$

and since $$z = ±√23$$ is not in the unit circle, it must be 0 right?

The second integral is

$$\oint \frac{\cos(z+11)}{z^2-6z} dz$$

which can be written as

$$\oint \frac{\cos(z+11)}{z(z-6)} dz$$

then

$$\oint \frac{\frac{\cos(z+11)}{z-6}}{z} dz$$

so evaluating $$\oint \frac{\cos(z+11)}{z-6} dz at z=0$$ gives $$\frac{pi*icos(11)}{3}$$

That's very good except the last one I get $-\frac{\pi i\cos(11)}{3}$ Also, you say "since $z=\pm \sqrt{23}$, really should say "the poles are at zero and $z=\pm \sqrt{23}$". What else? Yeah, that "dzat" thing. Kinda' problematic to include text and math code in a single line. Should have just put the integral, then say "at z" right below it. And the pi. The math code for that is \pi. Also, now that you posted that big notebook paper picture (any large pictures) it has skewed the entire thread so you can't view it without scrolling. That's a pain. Why not just remove it so it's easier to see the thread, then ask more question about it just using latex. Won't back here till tomorrow though. Others may help though.