Complex Analysis question
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Dr.Peterson
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No, it's wrong. It may be true that the image will be a subset of your annulus (or one like it), but there's a lot you haven't taken into account.
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].
Also, there are some specific errors, such as neglecting to square [imath]e^x[/imath].
Professor Steven,
Mapping circuits (disks) under [imath]e^z[/imath] always produces annulus. Therefore, from the last diagram, I think that the idea of the answer is correct. I am just lazy to check the correct numerical solution!
Mapping circuits (disks) under [imath]e^z[/imath] always produces annulus. Therefore, from the last diagram, I think that the idea of the answer is correct. I am just lazy to check the correct numerical solution!
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Dr.Peterson
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Really? I thought vertical lines mapped to circles, and horizontal lines to rays. So it would be vertical bars that would map to annuli, and that's really all @Steven G showed. (But since our disk is a subset of a vertical bar, its image will be a subset of that annulus.)
I'd like to see the source of the problem; it seems harder than the answer is worth (being hard to graph by hand), but then I haven't done complex analysis in a long time. Might it have actually asked for the pre-image? (In any case, it was copied incorrectly, with the word "image" twice:
Vertical and horizontal lines are mapped into circles and rays under [imath]\displaystyle w = \frac{1}{z}[/imath] while circles are mapped into annulus under [imath]w = e^z[/imath] with annulus radius equal to [imath]e^{-r} \leq |w| \leq e^r[/imath] where [imath]r[/imath] is the radius of the disk.
Dr.Peterson
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Let's check that claim, by mapping the four simplest points on the given circle:
[math]e^{\frac{\pi}{2}}\approx4.81\\e^{\frac{\pi}{2}i}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i\\e^{-\frac{\pi}{2}}\approx0.208\\e^{-\frac{\pi}{2}i}=\cos\frac{\pi}{2}-i\sin\frac{\pi}{2}=-i[/math]
These don't lie on a circle:
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Professor Dave, I might be wrong as I did not practise mapping for two years. But here is my analysis. The rotation of the disk is [imath]0 \leq \theta \leq 2\pi[/imath], so we have [imath]\displaystyle z = \frac{\pi}{2}(\cos \theta + i\sin\theta)[/imath]. This can be divided into [imath]\displaystyle x = \frac{\pi}{2}\cos \theta[/imath] and [imath]\displaystyle y = \frac{\pi}{2}\sin \theta[/imath] which are the real and imaginary parts respectively. Now, [imath]w = e^z = e^{x + iy} = e^{x}e^{iy} = e^x(\cos y + i\sin y)[/imath]. The magnitude of [imath]|w| = e^x[/imath], so we are interested in the real part from [imath]\displaystyle z[/imath] which is [imath]\displaystyle x = \frac{\pi}{2}\cos \theta[/imath].
The maximum of [imath]x[/imath] happens when [imath]\displaystyle x = \frac{\pi}{2}\cos 0 = \frac{\pi}{2}[/imath].
The minimum of [imath]x[/imath] happens when [imath]\displaystyle x = \frac{\pi}{2}\cos \pi = -\frac{\pi}{2}[/imath].
Then,
[imath]\displaystyle e^{-\frac{\pi}{2}} \leq |w| \leq e^{\frac{\pi}{2}}[/imath].
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I don't understand why you want to use [imath]\displaystyle w = e^{\frac{\pi}{2}i}[/imath] when [imath]\displaystyle w = e^{\frac{\pi}{2} + yi}[/imath].
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blamocur
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blamocur
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Your last inequality shows that the resulting line fits inside an annulus (as mentioned by @Dr.Peterson earlier), but it does not make the line itself being equal to that annulus.Very beautiful.
Professor blamocur, where did I go wrong in my analysis?
Do you mean this? [imath]\displaystyle e^{-\frac{\pi}{2}} \leq |w| \leq e^{\frac{\pi}{2}}[/imath].
How to correct it?
Dr.Peterson
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Why are you only interested in the magnitude of w?
These tell you only that the magnitude of w is greatest and least at the first and third points I listed.
Why do you think x is fixed? We're mapping any point on the circle, and both x and y change as you move around it.
I listed the points at E, N, W, and S positions on the preimage (circle), and mapped them all. The north and south positions are [imath]\frac{\pi}{2}i[/imath] and [imath]-\frac{\pi}{2}i[/imath], so I mapped those.
What you're mapping is the vertical line tangent to the circle on the right.
The square of which that is the right side maps to half of the annulus.
That would be the vertical bar containing the circle.
To correct it, you can do what I suggested in post 2, after correcting a typo:
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.
blamocur
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Yes. It is correct but not enough to desribe the curve.
Thanks a lot professor Dave for your explanation. I am sorry I feel rusty to answer your questions!
And thanks a lot professor blamocur.
I just involved myself in this discussion because I studied the subject a long time ago. It is fully proved that I missed a lot of ideas!
Professor Dave, how do I get the same as your drawing?
Look at my diagram. I used Desmos with this formula: [imath]\displaystyle r = e^{\frac{\pi}{2}(\cos \theta + \text{i}\sin \theta)}[/imath], fixing [imath]\text{i}[/imath] at zero. If change the value of [imath]\text{i}[/imath] to any value other than zero, the blue drawing rotates upside down.
Dr.Peterson
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That's just a polar equation, unrelated to what we've been discussing. The reason it doesn't work is that theta here is the angle to the point you're drawing, not the angle in the preimage.
I don't think Desmos understands complex numbers at all (though I could be wrong). Clearly if you can change the value of "i" to any real number, it can't be what you intend!
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.)
Aren't [imath]x[/imath] and [imath]y[/imath] just [imath]\sin \theta[/imath] and [imath]\cos \theta[/imath]?
Am I just going to say: [imath]z(t) = x(t) + \text{i}y(t)[/imath], then [imath] x(t) = \frac{\pi}{2}\cos t[/imath] and [imath] y(t) = \frac{\pi}{2}\sin t[/imath].
[imath]w(t) = e^{z(t)} = e^{x(t) + \text{i}y(t)}[/imath]
[imath]\displaystyle \boldsymbol{\tau}(t) = \left(e^{\frac{\pi}{2}\cos t}, \cos\left[\frac{\pi}{2}\sin t\right] + \text{i}\sin\left[\frac{\pi}{2}\sin t\right]\right)[/imath]
Dr.Peterson
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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]?
[imath]e^{x(t)} = e^{\frac{\pi}{2}\cos t}[/imath]
[imath]\displaystyle e^{\text{i}y(t)} = \cos y(t) + \text{i}\sin y(t) = \cos \left[\frac{\pi}{2}\sin t\right] + \text{i}\sin \left[\frac{\pi}{2}\sin t\right][/imath]
So
The real part of [imath]e^{x(t) + \text{i}y(t)}[/imath] is [imath]\displaystyle \cos \left[\frac{\pi}{2}\sin t\right]e^{\frac{\pi}{2}\cos t}[/imath]
And
The imaginary part of [imath]e^{x(t) + \text{i}y(t)}[/imath] is [imath]\displaystyle \sin \left[\frac{\pi}{2}\sin t\right]e^{\frac{\pi}{2}\cos t}[/imath]
Let me guess. Does this mean, the parametric curve should be written like this:
[imath]\displaystyle \boldsymbol{\tau}(t) = \left(\cos \left[\frac{\pi}{2}\sin t\right]e^{\frac{\pi}{2}\cos t}, \sin \left[\frac{\pi}{2}\sin t\right]e^{\frac{\pi}{2}\cos t}\right)[/imath]
???
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