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

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

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

Which would give us:

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)