Answer
\[e^{(3+4i)x}=e^{3x}\cos 4x+i\:e^{3x}\sin 4x\]
Work Step by Step
We will use Euler's formula
\[e^{i\theta}=\cos\theta+i\sin\theta\]
$e^{(3+4i)x}=e^{3x}e^{4ix}$
Using Euler's formula
$e^{(3+4i)x}=e^{3x}\left[\cos 4x+i\sin 4x\right]$
$e^{(3+4i)x}=e^{3x}\cos 4x+ie^{3x}\sin 4x$
Here,
$u(x)=e^{3x}\cos 4x\;\;$ and $\;\;v(x)=e^{3x}\sin 4x$
Hence,
$e^{(3+4i)x}=e^{3x}\cos 4x+ie^{3x}\sin 4x$.
