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 )
