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

[Mayhem]

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.

 

[Fred Wright]

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,
$$
I=5^8z_0\int_0^{1} [(z_0t + i]^7dt=5^8z_0\sum_{n=0}^{7}\int_0^{1}\begin{pmatrix}
7\\
n
\end{pmatrix} z_0^{n}t^{n} (i)^{7-n}dt
$$