Putting z = i z=i z = i in the following term z 4 + z 3 + z − 1 z^{4}+z^{3}+z-1 z 4 + z 3 + z − 1 , we have
i 4 + i 3 + i − 1 = 1 − i + i − 1 = 0 i^{4}+i^{3}+i-1=1-i+i-1=0 i 4 + i 3 + i − 1 = 1 − i + i − 1 = 0
thus z = i z=i z = i is a root of the equation z 4 + z 3 + z − 1 = 0 z^{4}+z^{3}+z-1=0 z 4 + z 3 + z − 1 = 0
Since the equation z 4 + z 3 + z − 1 = 0 z^{4}+z^{3}+z-1=0 z 4 + z 3 + z − 1 = 0 has the root z = i z=i z = i , then by using the division rule, we have that
z 4 + z 3 + z − 1 = ( z − i ) ( z 3 + ( i + 1 ) z 2 + ( i − 1 ) z − i ) z^{4}+z^{3}+z-1=(z-i)(z^{3}+(i+1)z^{2}+(i-1)z-i) z 4 + z 3 + z − 1 = ( z − i ) ( z 3 + ( i + 1 ) z 2 + ( i − 1 ) z − i )
Since z = i z=i z = i is a root of the equation z 4 + z 3 + z − 1 = 0 z^{4}+z^{3}+z-1=0 z 4 + z 3 + z − 1 = 0 , then z = − i z=-i z = − i is also a root, thus
z 3 + ( i + 1 ) z 2 + ( i − 1 ) z − i = ( z + i ) ( z 2 + z − 1 ) z^{3}+(i+1)z^{2}+(i-1)z-i=(z+i)(z^{2}+z-1) z 3 + ( i + 1 ) z 2 + ( i − 1 ) z − i = ( z + i ) ( z 2 + z − 1 )
the equation z 2 + z − 1 = 0 z^{2}+z-1=0 z 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 ) / 2 z= (-1+\sqrt5 )/2 z = ( − 1 + 5 ) /2 and z = ( − 1 − 5 ) / 2 z= (-1-\sqrt5 )/2 z = ( − 1 − 5 ) /2
the four roots are
z = i z=i z = i , z = − i z=-i z = − i , z = ( − 1 + 5 ) / 2 z= (-1+\sqrt5 )/2 z = ( − 1 + 5 ) /2 and z = ( − 1 − 5 ) / 2 z= (-1-\sqrt5 )/2 z = ( − 1 − 5 ) /2
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