1. ## line integral

$\int_C (z*^2dz + z^2dz*)$, where C is the curve defined by $z^2+2|z|^2+z*^2=2(1-i)z+2(1+i)z*$
$z$ is a complex number x+iy, $z*$ is his complex conjugate
Thanks

2. Originally Posted by hurz
$\int_C (z*^2dz + z^2dz*)$, where C is the curve defined by $z^2+2|z|^2+z*^2=2(1-i)z+2(1+i)z*$
$z$ is a complex number x+iy, $z*$ is his complex conjugate
Thanks
I think that there is a problem with your curve.

first I will use $z*=\bar{z}$ for complex conjugation
recall that
$|z|^2=z\bar{z}$ so we can factor the left hand side as follows

$z^2+z\bar{z}+z\bar{z}+\bar{z}^2=z(z+\bar{z})+\bar{ z}(z+\bar{z})=(z+\bar{z})^2=(2\text{Re}(z))^2=4x^2$

Now for the right hand side we get

$2(1-i)(x+iy)+2(1+i)(x-iy)=2[(x+iy-ix+y)+(x-iy+ix+y)]=4x+4y$

This gives the curve as

$4x^2=4x+4y \iff y=x^2-x$

If you expand out the integrand in a similar fashion you will get

$(2x^2-2y^2)dx+4xydy$
remember that
$dz=dx+idy; \quad d\bar{z}=dx-idy$
So this gives the real line integral

$\oint (2x^2-2y^2)dx+4xydy$
but the curve of integration (the parabola) is unbounded so the integral will be as well.

3. Originally Posted by TheEmptySet
I think that there is a problem with your curve.

first I will use $z*=\bar{z}$ for complex conjugation
recall that
$|z|^2=z\bar{z}$ so we can factor the left hand side as follows

$z^2+z\bar{z}+z\bar{z}+\bar{z}^2=z(z+\bar{z})+\bar{ z}(z+\bar{z})=(z+\bar{z})^2=(2\text{Re}(z))^2=4x^2$

Now for the right hand side we get

$2(1-i)(x+iy)+2(1+i)(x-iy)=2[(x+iy-ix+y)+(x-iy+ix+y)]=4x+4y$

This gives the curve as

$4x^2=4x+4y \iff y=x^2-x$

If you expand out the integrand in a similar fashion you will get

$(2x^2-2y^2)dx+4xydy$
remember that
$dz=dx+idy; \quad d\bar{z}=dx-idy$
So this gives the real line integral

$\oint (2x^2-2y^2)dx+4xydy$
but the curve of integration (the parabola) is unbounded so the integral will be as well.
If we evaluate the integral from z=1 to z=2(1+i), can i replace y for $x^2-x$ and dy for $(2x-1)dx$ in the integral and evaluate it from x=1 to x=2?

4. Originally Posted by hurz
If we evaluate the integral from z=1 to z=2(1+i), can i replace y for $x^2-x$ and dy for $(2x-1)dx$ in the integral and evaluate it from x=1 to x=2?
Yes that is correct

5. Originally Posted by TheEmptySet
Yes that is correct
Thanks!