Can any help me on where I should start to find all 8 solutions to x^8=3-4I . And also not exactly sure how to find the magnitude of each solution .
I think I got it but not sure yet. This problem is for a math 142 class, and the way you show is different from anything we have done in class. To find complex roots of something like 3-4I we would first write in polar form z=x+yi =r[cos(θ)+isin(θ)] . I would first find r=√(x^2+y^2) which equals 5. Then find θ since x is positive and y is negative it is in quadrant 3 sin(θ)=y/r =-4/5 . this is where im stuck because im not sure how to find θ in quad 3 whose sin is -4/5. do i convert -4/5 into degree by multiply by 180/π. I also know the magnitude is the same for all 8 solutions but i dont know how to.prove it with a complex in polar form.isnt the magnitude of a complex z=x-iy just √(x^2+y^2) which is 5 in this case. How would i find magnitude for each.solution if they are in polar form.
Sorry I meant 4 brain fart I guess. I have lost many hours of sleep over this. Im glad i found this forum. I couldn't find a tutor on campus that could help. If anything confused me even more. I'm going to do it again thanks for the help .