A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart.
One pair of sides is parallel to the imaginary axis.
R is the region outside the hexagon, and S = {1/z| z ε R}.
Given that:
Area of S has the form M*pi + √N, where M and N are positive integers.
Find M and N.
In addition to the hexagon, draw its incircle (diameter 1 given) and circumcircle. In polar coordinates those circles are r=1/2 and r=1/sqrt(3). The side intersecting the positive x-axis is the line 1/2 = r*cos(t) for -30deg <= t <= 30deg.
Let z be polar coordinate (r, t). Then the transform z -> 1/z becomes (r,t) -> (1/r, -t).
Then the r=1/2 circle becomes r=2, the r=1/sqrt(3) becomes r=sqrt(3), and the first side becomes r = 2*cos(t) for -30deg <= t <= 30deg. The last equation is a circle of radius 1, and the defined arc is a 120 degree sector of that circle, tangent to the r=2 circle at t=0 and intersecting the r=1/sqrt(3) at t = +/-30deg.
Six copies of this arc make up the transformation of the original hexagon. The chord spanning the two endpoints has a length sqrt(3). S is this six arc-lobed figure.
To help find the area of S, first draw the six chords separating S into a hexagon plus 6 circular segments. Divide the hexagon into six equilateral triangles. 3 of the circular segments plus one triangle makes one circle of radius 1.
Then the area of S equals the area of two circles of radius 1 plus 4 equilateral triangles of edge sqrt(3). Each circle has area of pi and each triangle has area 3*sqrt(3)/4. Then the area of S is 2*pi + 3*sqrt(3).
Solution
