Can the contour integral of z⁷ be simplified using a parameterized expression?

In summary, the conversation discusses parameterizing ##z## by ##z(t) = 5i + (3 + i - 5i)t## with given initial and final values, and rewriting the contour integral in terms of ##t##. The integral is then simplified using the binomial theorem and binomial coefficients.
  • #1
Mayhem
322
213
Homework Statement
Integrate ##\int_{\Gamma} z^7 dz## along the line ##5i## to ##3 + i##
Relevant Equations
##z = x + iy##
First I parameterize ##z## by ##z(t) = 5i + (3 + i - 5i)t## such that ##z(0) = 5i## and ##z(1) = 3 + i##, which means that ##0 \leq t \leq 0## traces the entire line on the complex plane. By distributing ##t##, we achieve a parameterized expression of the form ##z(t) = x(t) + iy(t)##
$$z(t) = 3t + i(5 - 4t)$$

Then to rewrite the contour integral in terms of ##t##, we determine ##dz/dt##
$$\frac{dz}{dt} = 3 - 4i$$
Yielding
$$\int_{\Gamma} z^7 dz = \int_{0}^{1} z^7 \frac{dz}{dt} dt = (3-4i)\int_{0}^{1} (3t + i(4-5t))^7 dt$$

Not sure where to go from here. The naive approach would be to expand the binominal and factor out all constants (including imaginary units ##i##), but that is tedious. Is there a trick? If this was a calc I problem, I'd just do a simple u-substitution, but not sure if the same logic holds when ##i## is involved.
 
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  • #2
Observe that you can rewrite your integral,
$$
I=(3-4i)\int_0^{1} [(3-4i)t + 5i]^7dt
$$
$$
(3-4i)=5e^{-i\tan^{-1}(\frac{4}{3})}=5z_0
$$
Expand the integral with binomial thm,
$$
I=5^8z_0\int_0^{1} [(z_0t + i]^7dt=5^8z_0\sum_{n=0}^{7}\int_0^{1}\begin{pmatrix}
7\\
n
\end{pmatrix} z_0^{n}t^{n} (i)^{7-n}dt
$$
 

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