If a,b,c are distinct integers and #omega# is a cube root of unity then minimum value of #|a + bomega + comega^2| + |a + bomega^2 + comega|#?

#|a + bomega + comega^2| + |a + bomega^2 + comega|#

1 Answer
George C.
Jan 21, 2018

#2sqrt(3)#

Explanation:

Note that given:

#omega = -1/2+sqrt(3)/2i#

#omega^2 = bar(omega) = -1/2-sqrt(3)/2i#

So:

#abs(a+bomega+comega^2) = abs((a-1/2b-1/2c)+sqrt(3)/2(b-c)i)#

#color(white)(abs(a+bomega+comega^2)) = sqrt(1/4(2a-b-c)^2+3/4(b-c)^2)#

#color(white)(abs(a+bomega+comega^2)) = 1/2sqrt((2a-b-c)^2+3(b-c)^2)#

Hence:

#abs(a+bomega+comega^2) + abs(a+bomega^2+comega) = sqrt((2a-b-c)^2+3(b-c)^2)#

In order to minimise this expression we need #a, b# and #c# to be close in value to one another.

Let's try:

#{ (b=-1), (a=0), (c=1) :}#

Then:

#sqrt((2a-b-c)^2+3(b-c)^2) = sqrt(0+12) = 2sqrt(3)#

Let's also try:

#{ (a=0), (b=1), (c=2) :}#

Then:

#sqrt((2a-b-c)^2+3(b-c)^2) = sqrt(9+3) = 2sqrt(3)#

Any other combination of distinct integers that we might try is either equivalent to one of these, or results in larger values.

So the minimum is #2sqrt(3)#

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