Question

if z-1/z =y then z⁴+1/z⁴ =​

Answer 1

Answer:

z 4 =(z−1) 4

⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0

⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2]=0

⇒(2z 2 −2z+1)(2z−1)=0

⇒z= 21 , 21 ± 21

i

Answer 2

Step-by-step explanation:

z

z 4

z 4 =(z−1)

z 4 =(z−1) 4

z 4 =(z−1) 4

z 4 =(z−1) 4 ⇒(z

z 4 =(z−1) 4 ⇒(z 2

z 4 =(z−1) 4 ⇒(z 2 )

z 4 =(z−1) 4 ⇒(z 2 ) 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1)

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ]

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1)

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1)

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z=

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 ,

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21 ±

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21 ± 2

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21 ± 21

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21 ± 21

z 4 =(z−1) 4 ⇒(z 2 ) 2 −[(z−1) 2 ] 2 =0⇒[z 2 +(z−1) 2 ][z 2 −(z−1) 2 ]=0⇒(2z 2 −2z+1)(2z−1)=0⇒z= 21 , 21 ± 21 i