What is the Correct Way to Find the Length of a Contour Gamma?

  • Thread starter Thread starter stunner5000pt
  • Start date Start date
  • Tags Tags
    Length
stunner5000pt
Messages
1,443
Reaction score
4
Find the length of the contour Gamma parametrized by z = z(t) = 5e^{3it}

the legnth of contour is \int_{0}^{\pi} \left|\frac{dz}{dt}\right| dt

now \left|\frac{dz}{dt}\right| = |15ie^{3it}| = |15ie^{3t}\cos(t) - 15e^{3t}\sin(t)| = \sqrt{225e^{6t} + 225e^{6t}}

is this right so far?? I think I am making a mess of the magnitude part
 
Physics news on Phys.org
stunner5000pt said:
Find the length of the contour Gamma parametrized by z = z(t) = 5e^{3it}

the legnth of contour is \int_{0}^{\pi} \left|\frac{dz}{dt}\right| dt

now \left|\frac{dz}{dt}\right| = |15ie^{3it}| = |15ie^{3t}\cos(t) - 15e^{3t}\sin(t)| = \sqrt{225e^{6t} + 225e^{6t}}

is this right so far?? I think I am making a mess of the magnitude part

The magnitude of 15ie^{3it} is 15! (just take the expression times its complex conjugate and then take the square root!).

Your second step is all wrong. You seem to think that e^{it} = e^t e^i ?? That`s of course incorrect!

Euler`s identity applied to e^{3 i t}gives cos(3t) + i sin(3t)

EDIT: which can also be written as (e^{it})^3 = (cos t + i sin t)^3 of course
 
Last edited:
nrqed said:
The magnitude of 15ie^{3it} is 15! (just take the expression times its complex conjugate and then take the square root!).

Your second step is all wrong. You seem to think that e^{it} = e^t e^i ?? That`s of course incorrect!

Euler`s identity applied to e^{3 i t}gives cos(3t) + i sin(3t)

EDIT: which can also be written as (e^{it})^3 = (cos t + i sin t)^3 of course

dont i feel stupid!
i was using Euler's identitiy incorrectly, when my course... for the most part... is based on it!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top