A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below:

Then what is \(\operatorname{Im}(a)\)?

Guest Jan 16, 2019

#1**0 **

if you tell me what Im(a) means(does it mean imaginary part?), i can solve it, but i can tell you the coordinates.

convert to the real plane, so distance from a to origin is sqrt2/2. now since the angle made by a, origin, and y axis is 15 degrees, you use the fact that cos 15 = (sqrt6+sqrt2)/4, and the fact that sin 15 = (sqrt6-sqrt2)/4, so then the coordinates of a would be ((sqrt6-sqrt2)/8, (sqrt6+sqrt2)/8).

assuming it means imaginary part, answer is (sqrt6+sqrt2)/8.

asdf335 Jan 16, 2019

#2**+9 **

**A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below:**

Then what is \(\operatorname{Im}(a)\)?

\(\begin{array}{|rcll|} \hline \sin(75^{\circ}) &=& \dfrac{\operatorname{Im}(a)} {\sqrt{2}} \\ \operatorname{Im}(a) &=& \sqrt{2}\cdot \sin(75^{\circ}) \\ \operatorname{Im}(a) &=& 1.41421356237\cdot 0.96592582629 \\ \mathbf{\operatorname{Im}(a)} & \mathbf{=} & \mathbf{1.36602540378} \\ \hline \end{array}\)

heureka Jan 16, 2019