Question

Find all complex solutions of the equation z⁴-1-i=0.​

Answer 1

Answer:

Step-by-step explanation:

hello :

z⁴=1+i      given : z² = t    t is the complex number  so :   z⁴= (z²)² =t²

solve for  t  this equation : t² = 1+i

if : t = x+iy     t² =(x+iy)² = x²+2xiy+(iy)² =x²-y²+2xyi ......     (  i² = -1)

t² = 1+i  means : x²-y²+2xyi = 1+i

you have this system :   x²-y² = 1.....(*)

                                      2xy  = 1.....(**)

slve for   x  and y    you ca add this equation :   ....(***)

( use : t² = 1+i

/t²/ = /1+i/   so :/t/² = /1+i/ .... (√(x²+y²))²=√(1²+1²) =√2  means x²+y² = √2)

the system is :   x²-y² = 1.....(*)

                         2xy  = 1.....(**)

                         x²+y² = √2 ....(***)

add (*) and (***) : 2x²= 1+√2

x² = (√2+1)/2

substrac  (*) and (***)  you have : y² = (√2-1)/2

use (**) the proudect  xy is positif (same sign)  so :

x=√( (√2+1)/2)  and  y = √( (√2-1)/2)   so :  t1 =√( (√2+1)/2)+i√( (√2-1)/2)  

x= -√( (√2+1)/2)  and  y = -√( (√2-1)/2) so :t2= -t1

same method for equation : z² =t1  ..(2 solution )  and z² = t2..(2 solution )