A and B beeing two complex, then A^4 = B^4 is equivalent to A = B.u with u a 4th root of 1, i.e. u = 1 or u = i or u = -1 or u = -i So here, with A = x - 3 and B = 2, the given equation is equivalent to x - 3 = 2.1 or x - 3 = 2.i or x - 3 = 2.(-1) or x - 3 = 2.(-i), then it is equivalent to x = 5 or x = 3 + 2.i or x = x = 1 or x = 3 - 2.i
Well done but a bit complex :-) an easiest way from my point of view would be to write x-3 = Z (complex). on the right side: Module 16 =16 , argument =0+2*k*pi, on the left side: Module Z^4 =16; argument Z^4 =4* alpha -> Module Z = 2 ; argument Z=k*Pi/2 with k=0,1,2,3 alpha =0, pi/2 , pi , 3*pi/2 so Z = 2(1+0*i) ; Z= 2*(0+i); Z=2*(-1+0*i); z= 2*(0-i) back to the initial variable x=z+3 so S= {1,5,3+2i, 3-2i}
A and B beeing two complex, then A^4 = B^4 is equivalent to A = B.u with u a 4th root of 1, i.e. u = 1 or u = i or u = -1 or u = -i
So here, with A = x - 3 and B = 2, the given equation is equivalent to x - 3 = 2.1 or x - 3 = 2.i or x - 3 = 2.(-1) or x - 3 = 2.(-i),
then it is equivalent to x = 5 or x = 3 + 2.i or x = x = 1 or x = 3 - 2.i
Very good!👏👏👏
Well done but a bit complex :-) an easiest way from my point of view would be to write x-3 = Z (complex).
on the right side: Module 16 =16 , argument =0+2*k*pi,
on the left side: Module Z^4 =16; argument Z^4 =4* alpha
-> Module Z = 2 ; argument Z=k*Pi/2
with k=0,1,2,3 alpha =0, pi/2 , pi , 3*pi/2 so Z = 2(1+0*i) ; Z= 2*(0+i); Z=2*(-1+0*i); z= 2*(0-i)
back to the initial variable x=z+3 so S= {1,5,3+2i, 3-2i}
X =5
5&minus 1
x=5 or 1
X=5 ? Dans R...et X=1