Quote Originally Posted by Plato View Post
There is no $\displaystyle i$ in the imaginary part.

$\displaystyle \text{Im}(a+bi)=b$ NOT $\displaystyle bi$. I would mark that wrong.
Ah yes, so it's:

So it's
$\displaystyle Im((a+bi)(c+di))=bc+ad$

Which would give us:

$\displaystyle Im\left( \frac{-3+4i}{25}(\cos(t)+i\cdot\sin(t)) \right)=Im\left( \left( \frac{-3}{25} + \frac{4}{25} \right) \left( \cos(t) + \cdot \sin(t)\right) \right)=\frac{4}{25}\cdot \cos(t) + \frac{-3}{25}\cdot \sin(t)$