I am trying to solve
$$y'' + 7y' + 10y = -4\sin(3t)$$
via the complex method of undetermined coefficients. I have already solved it via the trigonometric method and obtained $$y = \frac{42\cos(3t) - 2\sin(3t)}{221}.$$
The ODE is equivalent to $y'' + 7y' + 10y = -4\text{Im}(e^{3ti})$. Since $Im()$ is homogeneous of degree $1$, we should be able to simply move the constants obtained taking exponential derivatives outside of the $\text{Im}()$; thus the guess $y_p = a\text{Im}(e^{3ti})$ is a form replicated under differentiation. Plugging this guess into $$y'' + 7y' + 10y = -4\text{Im}(e^{3ti})$$ yields $$(a + 21ai)\text{Im}(e^{3ti}) = -4\text{Im}(e^{3ti}),$$ so $a = \frac{42i - 2}{221}$. This means $y_p = \frac{42\sin(3t)i - 2\sin(3t)}{221}$, which is not quite correct, as $\sin(\theta)i$ is not the the same as $\cos(\theta)$. What went wrong?
For the complex approach we should find a complex function $z_p$ such that $$z_p'' + 7z_p' + 10z_p = -4e^{3ti}.$$ Then, by letting $y_p=\text{Im}(z_p)$, we get $$y_p'' + 7y_p' + 10y_p = \text{Im}(-4e^{3ti})=-4\sin(3t).$$ We have that $z_p(t)=ae^{3ti}$ for some $a\in\mathbb{C}$ and therefore $$-9ae^{3ti}+21iae^{3ti}+10ae^{3ti}=-4e^{3ti}\implies a=\frac{-4}{1+21i}=\frac{-4}{1+21i}=\frac{-2+42i}{221}$$ (the same of your computation). Hence $$y_p=\text{Im}(z_p)=\text{Im}\left(\frac{-2+42i}{221}\cdot (\cos(3t)+i\sin(3t))\right) =\frac{42\cos(3t)-2\sin(3t)}{221}$$ which is exactly what you obtained by using the real approach.