Complex Analysis question

[mario99]

Close, but not quite. You have most of the pieces, but not assembled correctly.

What are the real and imaginary parts of [imath]e^{x(t) + \text{i}y(t)}[/imath], if [imath] x(t) = \frac{\pi}{2}\cos t[/imath] and [imath] y(t) = \frac{\pi}{2}\sin t[/imath]?

You are awesome professor Dave.



I know that parametrization is very basic, but the way how you noticed that the annulus sketch of the OP was wrong is super. I have always thought that transforming circles under [imath]e^z[/imath] gives annulus with inner radius [imath]e^{-r}[/imath] and outer radius [imath]e^r[/imath] where [imath]r[/imath] is the radius of the disk.

This makes me wondering what the heck the analysis of [imath]e^{-\frac{\pi}{2}} \leq |w| \leq e^{\frac{\pi}{2}}[/imath] even means????!!!

Professor Steven example in the OP was so beautiful and it showed how may someone get wrong easily when he thinks that he was always right. I hope that professor Steven caught his error and thanks a lot Dr.Peterson.

 

[Dr.Peterson]

This makes me wondering what the heck the analysis of [imath]e^{-\frac{\pi}{2}} \leq |w| \leq e^{\frac{\pi}{2}}[/imath] even means????!!!

That is the image of the infinite vertical strip [imath]-\frac{\pi}{2} \leq Re(z) \leq \frac{\pi}{2}[/imath].

1.8: Complex Functions as Mappings

math.libretexts.org

 

[mario99]

That is the image of the infinite vertical strip [imath]-\frac{\pi}{2} \leq Re(z) \leq \frac{\pi}{2}[/imath].

1.8: Complex Functions as Mappings

math.libretexts.org

Awesome.

My QuaDS (Quick and Dirty Script) shows a similar picture:

That doesn't really match mine, which I'm increasingly confident of. That may be because our problem doesn't start with a unit circle.

I think that you are right professor Dave about the sketch of professor blamocur that his quick and dirty script has done it on a unit circle.

I have done it on a unit circle and I got this result.

 

[blamocur]

That doesn't really match mine, which I'm increasingly confident of. That may be because our problem doesn't start with a unit circle.

I managed to miss the [imath]\frac{\pi}{2}[/imath] part and shown the image of a unit circle -- sorry. A more relevant plot, where the blue circle has radius [imath]\frac{\pi}{2}[/imath], seems to match yours better:

 

[blamocur]

I think that you are right professor Dave about the sketch of professor blamocur that his quick and dirty script has done it on a unit circle.

You are right.

 

[Dr.Peterson]

If anyone's interested, here's the original equation I plotted:

​

And here's the parametric version:

​

 

[mario99]

If anyone's interested, here's the original equation I plotted:

View attachment 38585​

And here's the parametric version:

View attachment 38586​

I am interested in the derivation of the first one if you do not mind!

 

[Dr.Peterson]

I am interested in the derivation of the first one if you do not mind!

Did you try doing what I said?

I would solve for x and y in terms of u and v, then set [imath]x^2+y^2\le\frac{\pi}{2}[/imath].

That's the derivation! Solve this for x and y, and then plug that into the equation of the disk:

​

Then, of course, change the variable names in order to graph it.

 

[mario99]

Did you try doing what I said?

I would solve for x and y in terms of u and v, then set [imath]x^2+y^2\le\frac{\pi}{2}[/imath].

Oops, I don't know how I missed that!

Then, of course, change the variable names in order to graph it.

That was a lot easier than the parametrization. But I think that you have a typo here: You meant to write [imath]u^2 + v^2 = e^{2x}[/imath], aren't you?

 

[Dr.Peterson]

Oops, I don't know how I missed that!

That was a lot easier than the parametrization. But I think that you have a typo here: You meant to write [imath]u^2 + v^2 = e^{2x}[/imath], aren't you?

That's not my mistake; I just forgot to correct it when I quoted it this time. What I said originally included

Also, there are some specific errors, such as neglecting to square [imath]e^x[/imath].

Luckily, you didn't need to be told that.

 

[blamocur]

That was a lot easier than the parametrization. But I think that you have a typo here: You meant to write u2+v2=e2xu^2 + v^2 = e^{2x}u2+v2=e2x, aren't you?

and, similarly, [imath]x^2+y^2 \leq \left(\frac{\pi}{2}\right)^2[/imath]

 

[mario99]

That's not my mistake; I just forgot to correct it when I quoted it this time. What I said originally included

Luckily, you didn't need to be told that.

I admit it was my fault (not yours) as I did not read carefully what you have suggested (and corrected). I was more focused on why my analysis (post #8) was wrong (not really wrong but it would not draw the annulus I had in mind) and did not pay a lot of attention at your post #2 which was suggested to professor Steven.

You can make a parametric curve, but you'll have to do the complex arithmetic yourself to calculate x and y. (I didn't do that, but wrote an equation the way I said at the start. I have now done this, and it gives the same result, except that it can't be an inequality.)

Also here, I did not understand what you meant by that (above), but rather focused on the parametric curve. Now I understand that you meant [imath]u[/imath] and [imath]v[/imath] from your post #2.

and, similarly, [imath]x^2+y^2 \leq \left(\frac{\pi}{2}\right)^2[/imath]

Thanks professor blamocur. This one I spotted it when professor Dave analyzed my analysis in his post #13.


May be if I have paid more attention to post #2, I would not have had to go through all of these discussions. But if that happened I would also miss the fun part which was parametrization. So thanks to what happened because I have now two ways to solve the problem, and maybe three including my analysis in post #8. Also my mapping concept to such problems was corrected. Thanks a lot professor Dave.