 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Hexagon in Complex Plane (Posted on 2016-03-13) 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.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
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).

 Posted by Brian Smith on 2016-03-13 14:31:42 Please log in:

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