I have to compute the integral $\int \vert z\vert dz $ following the next paths:
1)through the radio vector of the point $z=2-i$
2)through the semicircumference $\vert z\vert=1$, $0\le argz\le \pi$ (the path begins at the point $z=1$)
3)through the semicircumference $\vert z\vert=1$, $-\frac{\pi}{2}\le argz\le \frac{\pi}{2}$ (the path begins at the point $z=-i$)
4)through circumference $\vert z\vert=R$
The solution for 4) I think is $\int_R \vert z\vert dz=0 $ because $\vert z\vert=R$ it's a closed Jordan curve.
for 3), I have this $\int_J{|z|}\mathrm{dz}=-\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}ie^{i\theta}d\theta=-i\frac{e^{i\theta}}{i}|_{\frac{\pi}{2}}^{\frac{3\pi}{2}}=-e^{\frac{i3\pi}{2}}+e^{\frac{i\pi}{2}}=-(-i)+i=2i$
For 2), I have this $\int_J{|z|}\mathrm{dz}=\int_{0}^{\pi}ie^{i\theta}d\theta=e^{i\pi}-1=-2$
And for 1) I don't know how to integrate. The curve is a line?
Can someone help me with 1)? Also are my others answers correct?
Thanks in advance
I'm not sure I understand your English, but if you are asking in (1) to integrate along the line segment from the origin to the point $2-i$, you should parametrize that path by $z(t) = t(2-i)$, $0\le t\le 1$. Now what is $$\int_C |z|dz = \int_0^1 |z(t)|z'(t)\,dt = \int_0^1 \sqrt 5t(2-i)\,dt?$$