Countour integral of z from 0 to 1+2i

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Homework Statement



evaluate contour integral ∫ Z dz from 0 to 1+21 in the curve of y=2x
 
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doey said:

Homework Statement



evaluate contour integral ∫ Z dz from 0 to 1+21 in the curve of y=2x


So what exactly is your problem here? I can think of two ways to do this. Parametrize the curve or use the complex antiderivative. Both are easy. Do something!
 
Last edited:
Dick said:
So what exactly is your problem here? I can think of two ways to do this. Parametrize the curve or use the complex antiderivative. Both are easy. Do something!

ok,i try to solve for integrate z dz= x+iy(dx+idy) and get dy dx from y=2x.and i get the answer for integrate z dz,
how about the answer for integrate complex conjugate of z? since complex conjugate of z is not analytical .can i do it in the same way oso?
 
Last edited:
doey said:
ok,i try to solve for integrate z dz= x+iy(dx+idy) and get dy dx from y=2x.and i get the answer for integrate z dz,
how about the answer for integrate complex conjugate of z? since complex conjugate of z is not analytical .can i do it in the same way oso?

Yes, you can do it in the same way.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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