What are the solutions to the equation (z+1)^4=1-i?

[matpo39]

find all solutions of the given equation: (z+1)^4=1-i
im not sure if i did this right, but here's what i did
the first thing that i did was notice that
1-i = 2^1/2 * (cos (pi/4) + i*sin(pi/4))
then i found
z= [2^(1/2*1/4) * (cos (pi/4) + i*sin (pi/4) )^1/4] -1
then using de moivre's thrm
z= {2^3/4 *(cos [pi/16 +2k*pi/16] + i*sin[pi/16 + 2k*pi/16])}-1 ; k=0,1,2,3
does this look right?
thanks

 

[d_leet]

find all solutions of the given equation: (z+1)^4=1-i
im not sure if i did this right, but here's what i did
the first thing that i did was notice that
1-i = 2^1/2 * (cos (pi/4) + i*sin(pi/4))
then i found
z= [2^(1/2*1/4) * (cos (pi/4) + i*sin (pi/4) )^1/4] -1
then using de moivre's thrm
z= {2^3/4 *(cos [pi/16 +2k*pi/16] + i*sin[pi/16 + 2k*pi/16])}-1 ; k=0,1,2,3
does this look right?
thanks

Well the angle isn't pi/4 for the first thing. The point corresponding to 1-i is (1, -1)

 

[matpo39]

whoops, my bad
1-i = (2^1/2) * [cos (-pi/4) + i*sin(-pi/4)]

If i use this for 1-i and use the method of the first post will i obtain the correct answer?

thanks

 

[Bacat]

I just worked it out using your method and it looks good to me. You end up with four roots and they seem to correspond numerically to the values that Mathematica produces.