We can obtain [tex]a[/tex] by multiplying [tex]1-2i[/tex] by [tex]e^{i2\pi/3}[/tex], since this corresponds to rotating the point [tex]1-2i[/tex] in the plane counterclockwise about the origin by an angle of [tex]\frac{2\pi}3[/tex], the measure of any interior angle of the hexagon.
[tex]a \, e^{i2\pi/3} = (1 - 2i) \, e^{i2\pi/3} \\\\ ~~~~~~~~= (1 - 2i) \left(\cos\left(\dfrac{2\pi}3\right) + i \sin\left(\dfrac{2\pi}3\right)\right) \\\\ ~~~~~~~~ = (1 - 2i) \left(-\dfrac12 + i\dfrac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \sqrt3-\dfrac12 + i\left(1 + \dfrac{\sqrt3}2\right)[/tex]
Then
[tex]\mathrm{Im}(a) = \boxed{1 + \dfrac{\sqrt3}2}[/tex]
