Complex integration Watch

Goldenratio
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#1
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#1
Hi

I'm stuck with \int_C |z|dz
around a) a straight line from -i to i
b) a segment of a circle from -i to i


my parametrised eq for a) is f(t)=-i+2it

Thank you
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DFranklin
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#2
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(Original post by Goldenratio)
Hi

I'm stuck with \int_C |z|dz
around a) a straight line from -i to i
b) a segment of a circle from -i to i


my parametrised eq for a) is f(t)=-i+2it
Then |z| = |2t-1|, which is going to be a bit awkward.

I suggest you parameterize from -i to 0 and 0 to i instead.
Integral on the bit from -i to 0, f(t)=(t-1)i, |z| = 1 - t, and dz/dt = i.

So you get \int_C |z| dz = \int_0^1 (1-t)i\, dt.

You then need to do a similar integral for the bit from 0 to i and add them together.

Does that make any sense?
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Goldenratio
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#3
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Thank you! that makes sense!

I'm still a bit stuck on how you found |z|=|2t-1|? or f(t)=(t-1)i, |z| = 1 - t

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DFranklin
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(Original post by Goldenratio)
Thank you! that makes sense!

I'm still a bit stuck on how you found |z|=|2t-1|? or f(t)=(t-1)i, |z| = 1 - t
Well, f(t) is just z, right? So if f(t) = -i+2it (your parameterization) then |z|=|f(t)| = |-i+2it| = |2t-1|.

So what I did for my parameterization is pretty much the same, only I parameterize z=-i+it so as to go from -i to 0 as t goes from 0 to 1.
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Goldenratio
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#5
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hmm It's finding the modulus of those parametrizations that i'm not quite understanding. Could you explain it a bit more please?
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DFranklin
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#6
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Suppose we're integrating along the straight line from -i to 0.

Parameterize it as z = f(t) = -i + it. (0<=t<=1)

Then |z| = |-i+t| = |-1+t| = |1-t| = 1-t

Meanwhile, dz/dt = i. So \int ... dz = \int ... dz/dt dt = \int ... i dt

So our integral becomes \int_0^1 (1-t)i \, dt.

I'm not sure this is really any clearer than what I said before, so if you still don't understand, can you specify the exact part you don't follow, and I'll see if I can do better.
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Goldenratio
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#7
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I was referring to finding the modulus of a parametrisation or any complex number not in the form z=x+iy.

As in our cases:

|z|=|i+it| or |e^it| or for example |it^2+t+i|

I'm not sure how i work it out, because i just thought it was the length of the curve.
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DFranklin
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(Original post by Goldenratio)
I was referring to finding the modulus of a parametrisation or any complex number not in the form z=x+iy.

As in our cases:

|z|=|i+it| or |e^it| or for example |it^2+t+i|

I'm not sure how i work it out, because i just thought it was the length of the curve.
It's not the length of the curve, it's the distance between z and the origin. (i.e. it's |z| = \sqrt{x^2+y^2} when z = x+iy).

So in the case of |-i+it| you can calculate the modulus as I did earlier, or you could, equally say you've parameterized along a straight line starting distance 1 and ending at the origin, so the modulus is just 1-t.

Similarly, for e^it you can either show the modulus is \sqrt{\sin^2 t + \cos ^2 t} = 1, or you can just say you're integrating along part of a circle centered at the origin with radius 1, so the modulus is 1.

I'm slightly concerned we're talking at cross-purposes, so if any of this comes across as stating the bleeding obvious, don't take it the wrong way!

Edit: for your example, |it^2+t+i| = |(t^2+1)i + t| = \sqrt{(t^2+1)^2+t^2} = \sqrt{t^4+3t^2+1} (not a very nice answer, I admit!).
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Goldenratio
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#9
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No no, that was quite helpful! thank you.

In this example

|it^2+t+i| = |(t^2+1)i + t| = \sqrt{(t^2+1)^2+t^2} = \sqrt{t^4+3t^2+1}

it seems quite logical.

yet when we're getting the modulus of |-i+it| we don't end up with \sqrt{(-1+t)^2} but rather (i)^2(-1+t) which i must be misunderstanding completely
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DFranklin
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#10
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(Original post by Goldenratio)
No no, that was quite helpful! thank you.

In this example

|it^2+t+i| = |(t^2+1)i + t| = \sqrt{(t^2+1)^2+t^2} = \sqrt{t^4+3t^2+1}

it seems quite logical.

yet when we're getting the modulus of |-i+it| we don't end up with \sqrt{(-1+t)^2} but rather (i)^2(-1+t) which i must be misunderstanding completely
You do end up with \sqrt{(-1+t)^2} , it's just that you need to make sure you take the positive square root. (And as -1+t < 0, that's not the obvious one).
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Goldenratio
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#11
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aha! excellent!

So now i'm parametrising the circle i've got:

|g(t)|=|e^{\frac{i\pi}{2} t}|=1

and g'(t)=i\pi/2 e^{\frac{\pi}{2} t}

hence \int_0 ^1 g(t)g' = \pi/4(1-i) which is wrong
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DFranklin
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#12
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You haven't made it obvious how you're parameterizing, but you do realise you need to go from e^{-i\pi/2} to e^{i\pi/2}?
(i.e. the argument changes by pi, not pi/2).

Incidentally, I think the answer will depend on whether you go clockwise or anticlockwise around the circle. (Edit: Actually, I think that's not true in this case, but it's something to be aware of).
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Goldenratio
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#13
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the Parameterisation I'm using is g(t)= e^i*pi/2*t where |g(t)|=1 and g'=i*pi/2*e^1*Pi/2*t

tE[0,1)
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DFranklin
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#14
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(Original post by Goldenratio)
the Parameterisation I'm using is g(t)= e^i*pi/2*t where |g(t)|=1 and g'=i*pi/2*e^1*Pi/2*t

tE[0,1)
Then I think your g only goes round a quarter of a circle (from 1 to i), rather than half a circle (from -i to i).
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