What are the roots of #Z^4-Z^3+Z^2-Z+1 = 0# ?

1 Answer
George C.
Nov 11, 2017

Roots:

#Z = 1/4(1+sqrt(5))+-1/4sqrt(10-2sqrt(5))i#

#Z = 1/4(1-sqrt(5))+-1/4sqrt(10+2sqrt(5))i#

Explanation:

Note that:

#(Z+1)(Z^4-Z^3+Z^2-Z+1) = Z^5+1#

So the zeros of #Z^4-Z^3+Z^2-Z+1# are the fifth roots of #-1# apart from #-1#.

They form #4# of the #5# vertices of a pentagon with unit radius in the complex plane...

graph{((x-cos(pi/5))^2+(y-sin(pi/5))^2-0.001)((x-cos(pi/5))^2+(y+sin(pi/5))^2-0.001)((x-cos(3pi/5))^2+(y-sin(3pi/5))^2-0.001)((x-cos(3pi/5))^2+(y+sin(3pi/5))^2-0.001) = 0 [-2.5, 2.5, -1.25, 1.25]}

In trigonometric form, they can be written:

#cos(pi/5)+-isin(pi/5)#

#cos((3pi)/5)+-isin((3pi)/5)#

This is a direct consequence of de Moivre's formula:

#(cos theta + i sin theta)^n = cos n theta + i sin n theta#

e.g.:

#(cos(pi/5)+isin(pi/5))^5 = cos pi + i sin pi = -1#

Alternatively, note that:

#(Z^2-(1/2+sqrt(5)/2)Z+1)(Z^2-(1/2-sqrt(5)/2)Z+1)#

#=Z^4-Z^3+Z^2-Z+1#

Note that:

#(1/2+sqrt(5)/2)^2-4 = (3/2+sqrt(5)/2)-4#

#color(white)((1/2+sqrt(5)/2)^2-4) = -1/4(10-2sqrt(5))#

Hence using the quadratic formula, we find that the zeros of the given quartic are:

#1/4(1+sqrt(5))+-1/4sqrt(10-2sqrt(5))i#

#1/4(1-sqrt(5))+-1/4sqrt(10+2sqrt(5))i#

For example:

#Z^2-(1/2+sqrt(5)/2)Z+1 = 0#

is in standard form:

#aZ^2+bZ+c = 0#

with #a=1#, #b=-(1/2+sqrt(5)/2)# and #c=1#.

So it has roots given by the quadratic formula:

#Z = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(Z) = ((1/2+sqrt(5)/2)+-sqrt(-1/4(10-2sqrt(5))))/2#

#color(white)(Z) = 1/4(1+sqrt(5))+-1/4sqrt(10-2sqrt(5))i#

Similarly, #Z^2-(1/2-sqrt(5)/2)Z+1 = 0# has roots:

#Z = 1/4(1-sqrt(5))+-1/4sqrt(10+2sqrt(5))i#

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