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 Complex Number Query (Posted on 2016-04-03)
Determine all possible values of a complex number z such that:
(3z+1)(4z+1)(6z+1)(12z+1) = 2

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution (probably the hard way) Comment 1 of 1
There's probably a more clever method than just multiply out, set to zero, and re-factor.  But it worked.  I'd like to see a better way, but this was a nice diversion.

The product is
864z^4+720z^3+210z^2+25z+1=2
set to zero
864z^4+720z^3+210z^2+25z-0=0

I figured if this is solvable its probably the product of two quadratics so I decided to try factoring by hand as
(az^2+bz+1)(dz^2+ez-1)=864z^4+720z^3+210z^2+25z-1
ae+bd=720
-a+be+d=210
-b+e=25

solving for b gives
b=(720a-25a^2)/(a^2+864)
which only has a few solutions with positive integer for a and only one of these gives positive integers for the rest.  The factorization sought is:

(12z^2+5z+1)(72z^2+30z-1)=0

and the quadratic formula finishes us off.  The solutions are:

(-5±i√23)/24 and (-5±√33)/24
 Posted by Jer on 2016-04-03 20:19:34

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