How do I compute $ \int _{C} (z^3 + 2z +{\bf Re} z)\,dz$ where C is a triangle of vertices $z=0$, $z=1+2i $ and $z=1$.
The solution given is $i$
Anyone showing me how to deal with these problems will be extremely helpful, as this entire subject quite unclear to me.
On
Alternatively, you say that $z=x+yi$ and rewrite the integral as $\oint_C\left[(x+yi)^3+2(x+yi)+x\right]\,\left(\mathrm{d}x+i\mathrm{d}y\right)$. This is a simple closed curve on the complex plane therefore we can use greens theorem (who’s complex equivalent I cannot remember) \begin{align*} \oint_C\left[(x+yi)^3+2(x+yi)+x\right]\,\mathrm{d}x+ \left[(x+yi)^3+2(x+yi)+x\right] i\mathrm{d}y&=\iint_Di\left(3(x+yi)^2+3\right)-\left(3i(x+yi)^2(i)+2i\right)\,\mathrm{d}A\\ &=i\iint_D 3(x+yi)^2+3-3(x+yi)^2-2\,\mathrm{d}A\\ &=i\iint_D\,\mathrm{d}A \end{align*} The triangular region we are integrating along is simply a right triangle with a height of 2 and base of 1 so the area is simply $\frac12(1)(2)=1$ Multiplying this by $i$ we get the final answer of $i$
The function $z^3+2z$ in triangle is analytic so it's integral over $C$ is zero. With paramerization of $C$ we have \begin{cases} C_1:~t+2it&0\leq t\leq1, \\ C_2:~1+2i(1-t)&0\leq t\leq1, \\ C_3:~1-t&0\leq t\leq1. \end{cases} this parametrization is counter-clockwise (according to the vertices in the question), so we split the integral over $C$: \begin{align} \int_{C} (z^3 + 2z +{\bf Re} z)dz &= \int_{C} {\bf Re} z\,dz \\ &= \int_{C_1} {\bf Re} z\,dz+\int_{C_2} {\bf Re} z\,dz+\int_{C_3} {\bf Re} z\,dz \\ &= \int_0^1t(1+2i)\,dt+\int_0^1-2i\,dt+\int_0^1(1-t)(-1)\,dt \\ &= \int_0^1(2t+2it-2i-1)\,dt \\ &= \color{blue}{-i} \end{align} if we change the direction of path, we will have $i$ as answer instead of $-i$.