Question #116894
Show that z = i is a root of the equation z
+ z + z − 1 = 0. Find the three
other roots.
1
Expert's answer
2020-05-25T21:11:57-0400

Putting z=iz=i in the following term z4+z3+z1z^{4}+z^{3}+z-1 , we have

i4+i3+i1=1i+i1=0i^{4}+i^{3}+i-1=1-i+i-1=0

thus z=iz=i is a root of the equation z4+z3+z1=0z^{4}+z^{3}+z-1=0

Since the equation z4+z3+z1=0z^{4}+z^{3}+z-1=0 has the root z=iz=i , then by using the division rule, we have that

z4+z3+z1=(zi)(z3+(i+1)z2+(i1)zi)z^{4}+z^{3}+z-1=(z-i)(z^{3}+(i+1)z^{2}+(i-1)z-i)

Since z=iz=i is a root of the equation z4+z3+z1=0z^{4}+z^{3}+z-1=0 , then z=iz=-i is also a root, thus


z3+(i+1)z2+(i1)zi=(z+i)(z2+z1)z^{3}+(i+1)z^{2}+(i-1)z-i=(z+i)(z^{2}+z-1)

the equation z2+z1=0z^{2}+z-1=0 can be solved by the general law of the quadratic function , the roots of this equation is are z=(1+5)/2z= (-1+\sqrt5 )/2 and z=(15)/2z= (-1-\sqrt5 )/2


the four roots are

z=iz=i , z=iz=-i , z=(1+5)/2z= (-1+\sqrt5 )/2 and z=(15)/2z= (-1-\sqrt5 )/2





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