Find all complex a,b that solve the equation. When I was solving a DE problem I was able to reduce it to 
$$e^x \sin(2x)=a\cdot e^{(1+2i)x}+b\cdot e^{(1−2i)x}.$$ 
For complex $a,b$. Getting one solution is easy $(\frac{1}{2i},-\frac{1}{2i})$ but I was wondering what are all the values for complex $a,b$ that satisfy the equation. 
 A: When $x=0$ we have $0=a+b.$ So $b=-a.$ So $e^x\sin 2x=ae^{(1+2i)x}-ae^{(1-2i)x}=2iae^x\sin 2x.$ When $\sin 2x\ne 0$ this reduces to $1=2ia.$
A: Going through it step by step without the use (of frankly very useful) shortcuts.
Start with: 
$$e^x \sin(2x)=a\cdot e^{(1+2i)x}+b\cdot e^{(1−2i)x}$$
Clean up a bit and divide be $e^x$ on both sides:
$$  e^x \sin(2x)=a\cdot e^{x} e^{2xi}+b\cdot e^{x} e^{-2xi}$$
$$\sin(2x)=a\cdot  e^{2xi}+b\cdot e^{-2xi}$$
Throw in Euler's Formula: $e^{ix} = \cos(x) + i \sin(x)$
$$\implies  \sin(2x)=a\cdot \left(\cos(2x) + i \sin(2x)\right)+b\cdot \left(\cos(-2x) + i \sin(-2x)\right)$$
Note that: $\cos(x) = \cos(-x)$ and that $\sin(a) = -\sin(-a)$
$$\implies  \sin(2x)=a\cdot \left(\cos(2x) + i \sin(2x)\right)+b\cdot \left(\cos(2x) - i \sin(2x)\right)$$
$$\implies  \sin(2x)= \left( a+b \right)\cos(2x) +\left(a-b\right) i \sin(2x)  $$
We thus left with
$$\begin{cases} a+ b = 0   \\ \left(a-b\right) i = 1\end{cases}$$
Solving I only get one solution (which is also complex).
$$\boxed{a = \frac{-i}{2}} \text{ and } \boxed{ b = \frac{i}{2}}$$
This is the same as your solution, so I conclude no other (complex or not) solutions exist.
