# Check solution of integral

• Apr 20th 2010, 06:41 AM
piglet
Check solution of integral
I'm preparing for an exam and would like someone to check this if possible

$\displaystyle \int_{C}(z^{2}+z)dz$ where $\displaystyle C$ is any path from $\displaystyle -i$ to $\displaystyle 2+i$

I figured out that $\displaystyle (z^{2}+z)$ was not analytic using cauchy riemann equations

So i parametrized $\displaystyle -i$ to $\displaystyle 2+i$ using the formula
$\displaystyle z(t) = a + (b-a)t$ where $\displaystyle 0<=t<=1$

Filling into the above formula i got $\displaystyle z(t) = 2t + i(t-1)$
and $\displaystyle \frac{dz}{dt} = 2+i$

So i said $\displaystyle \int_{C}(z^{2}+z)dz = \int^{1}_{0}[(2t + i(t-1))^{2} + (2t +i(t-1))](2+i)dt$

i proceed to work this out an got an answer of $\displaystyle -\frac{5}{3} - \frac{4}{3}i$

The answer isn't really that important to me, i just would like to know if the line above my answer is correct

Thanks,
Piglet
• Apr 20th 2010, 07:15 AM
Magus01
Shouldn't the parametrisation be 2t + i(2t - 1) instead, because yours never attains the point 2+i for t in [0,1]. Otherwise you have the right idea with regards to computing the integral.
• Apr 20th 2010, 07:21 AM
piglet
Quote:

Originally Posted by Magus01
Shouldn't the parametrisation be 2t + i(2t - 1) instead, because yours never attains the point 2+i for t in [0,1]. Otherwise you have the right idea with regards to computing the integral.

$\displaystyle \int_{C}(z^{2}+z)dz = \int^{-1}_{1}[-(t+1)+(2t^{2}+3t+1)i]dt=\int^{-1}_{1}(2t^{2}+3t+1)idt-2=(4/3+2)i-2$