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Suppose $\gamma$ is a line segment between 1 and $3 + i$. Calculate $\int_{\gamma}z^2dz$ by definition and using the fact that $\int_{\gamma}f(z)dz = F(b) - F(a)$.

Now the second task seems almost like integral over reals:

$$ \int_{\gamma}z^2dz = \frac{z^3}{3}\Big|_1^{3+i} = \frac{(3+i)^3 - 1}{3} $$

I got stuck on the first part though as I just got introducted to the complex integrals.

$$ \int_{\gamma}f(x)dz = \int_a^bf(\gamma(s))\gamma'(s)ds $$

How can I find the $\gamma$? Is it the line $y=\frac{1}{2}x - \frac{1}{2}$? How can I transfer that into complex numbers?

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If you want to find a line segment from $z_0$ to $\zeta_0$ you can consider $\gamma:[0,1]\rightarrow \mathbb{C}$ defined by

$$\gamma(t) = (1-t)z_0+t\zeta_0$$

In your case this becomes

$$\gamma(t) = (1-t)\cdot 1+t(3+i) = 1+2t+it.$$

$\gamma'(t) = 2+i$. Now insert this into your integral...

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  • $\begingroup$ thanks there were a many typos! But I think I fixed them $\endgroup$ – OgvRubin Mar 18 '20 at 21:23
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The line is $(1-s)(1)+s(3+i)$. You get $\int_0^1f(\gamma(s))(2+i)\operatorname ds=(2+i)\int_0^1(2s+1+is)^2\operatorname ds=(2+i)\int_0^1(4s^2+4s+(4s+2)is-s^2+1)\operatorname ds=(2+i)[(3+4i)/3s^3+2s^2+is^2+s]_0^1=(2+i)((3+4i)/3+2 +i+1)=(2+i)(4+7/3i)=17/3+26/3i$.

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