Expressing $2e^{i\pi/2}+4e^{4i\pi/3}$ in $a+bi$ form I am asked to express
$$2e^{i\pi/2}+4e^{4i\pi/3}$$ in the form $a + bi$.
If I use Euler's theorem, then $e^{iz}=\cos(z)+i \sin(z)$
which gives me $$2e^{i\pi/2}+4e^{4i\pi/3}=2\cos(\pi/2)+i2\sin(\pi/2)
+4\cos(4\pi/3)+i4\sin(4\pi/3) \tag1$$
which simplifies to
$$0+2i-2-2i\sqrt3 \tag2$$
which simplifies to
$$-2+2i-2i\sqrt3 \tag3$$
Is this correct, or could this be further simplified?

That's correct so far. The last step is to factor out an $i$ to get:
$$-2 + (2 - 2 \sqrt3) i$$
in the form $a + bi$.
