# compute the integral of a complex-valued polynomial along a curve

• Oct 8th 2013, 09:20 PM
chen09
compute the integral of a complex-valued polynomial along a curve
$\gamma \text{ is any curve joining } -1+i \text{ to }1.\\ \text{ Letting }\gamma(b)=1, \, \gamma(a)=-1+i \text{ and know that } (-1+i)^2 = -2i,$ \begin{align*} \int_\gamma (z^3-6z^2 +4)dz &= \frac{\gamma^4(b)-\gamma^4(a)}{4}-6\frac{\gamma^3(b)-\gamma^3(a)}{3}+4(\gamma(b)-\gamma(a)) \\&= \frac{1^4- (-1+i)^4}{4}-6\frac{1^3- (-1+i)^3}{3}+4(1 - (-1+i)) \\ &= \frac{1^4- (-2i)^2}{4}-2(1^3- (-1+i)^2(-1+i))+4(2- i)) \\ &= \frac{1-(-4i)}{4} -2(1- (-2i)(-1+i)) +8 - 4i \\ &= \frac{1+ 4i}{4} -2(-1 -2i) +8 - 4i\\ &= \frac{41}{4} +i \end{align*}

This question is taken from Fisher's "Complex Variables 2nd Edition" on page 74.

Is my computation correct? Let me know.

Thanks a lot.
• Oct 8th 2013, 10:34 PM
Prove It
Re: compute the integral of a complex-valued polynomial along a curve
I think there is a much easier parameterisation. If you are going from the point \displaystyle \begin{align*} z = 1 + 0i \end{align*} and going to \displaystyle \begin{align*} z = -1 + i \end{align*}, that means you are trying to find the line between the points (1, 0) and (-1, 1).

The direction vector is \displaystyle \begin{align*} (-1 - 1, 1 - 0) = (-2, 1) \end{align*}, and so your line is \displaystyle \begin{align*} (-2, 1)t + (1, 0) \end{align*}.

Therefore your parameterisation is \displaystyle \begin{align*} x = -2t + 1 \end{align*} and \displaystyle \begin{align*} y = t \end{align*} with \displaystyle \begin{align*} 0 \leq t \leq 1 \end{align*}. And therefore your line \displaystyle \begin{align*} \gamma \end{align*} is given by the function \displaystyle \begin{align*} z = x + i\,y = -2t + 1 + i\,t \end{align*}.

So evaluating your integral:

\displaystyle \begin{align*} \int_{\gamma}{z^3 - 6z^2 + 4\,dz} &= \int_0^1{\left( -2t + 1 + i\,t \right) ^3 - 6 \left( -2t + 1 + i\,t \right) ^2 + 4 \, dt} \end{align*}

Go from here.
• Oct 8th 2013, 11:00 PM
chen09
Re: compute the integral of a complex-valued polynomial along a curve
As I was using your method, it gave me the right answer and also, showed me where I did wrong on my previous solution (I had (-2i)^2 = -4i, which should be just -4).

Using the result of a theorem to evaluate this is more efficient, but I appreciate all the help man!
• Oct 9th 2013, 12:54 AM
Prove It
Re: compute the integral of a complex-valued polynomial along a curve
Just out of interest, which theorem is more efficient in this case?