Question

A regular hexagon has vertices at
`1-2i, 0, a, b, c, d` as in the picture.

What is the Imaginary part of
`a`?

Answer 1

We can obtain
`a` by multiplying
`1-2i` by
`e^(i2\pi/3)`, since this corresponds to rotating the point
`1-2i` in the plane counterclockwise about the origin by an angle of
`\frac{2\pi}3`, the measure of any interior angle of the hexagon.

`a \, e^(i2\pi/3) = (1 - 2i) \, e^(i2\pi/3) \\\\ ~~~~~~~~= (1 - 2i) \left(\cos\left(\frac{2\pi}3\right) + i \sin\left(\frac{2\pi}3\right)\right) \\\\ ~~~~~~~~ = (1 - 2i) \left(-\frac12 + i\frac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \sqrt3-\frac12 + i\left(1 + \frac{\sqrt3}2\right)`

Then

`\mathrm{Im}(a) = \boxed{1 + \frac{\sqrt3}2}`