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Find all values of the real number \(a\) so that the four complex roots of \(z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0\) form the vertices of a parallelogram in the complex plane. Enter all the values, separated by commas.

 Jun 6, 2019
 #1
avatar+5662 
+3

a=3 is the only answer I see.

 Jun 7, 2019
 #2
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Thanks Rom but how did you find it ?

(Melody)

 Jun 7, 2019
edited by Melody  Jun 7, 2019
edited by Melody  Jun 7, 2019
 #3
avatar+5662 
+3

well....

 

I coded up a Mathematica sheet that plotted the 4 roots on the complex plane as I adjusted the parameter a.

 

Having seen it was right about 3 I then checked the roots.

 

Unless there's some trick this is an absurdly difficult problem w/o software.

Rom  Jun 7, 2019
 #4
avatar+102791 
0

ok thanks Rom   wink

Melody  Jun 9, 2019
 #5
avatar+140 
+1

Thank you so much! :)

 Jun 12, 2019

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