Dayv O. answered 02/26/21
Caring Super Enthusiastic Knowledgeable Trigonometry Tutor
Zoe, I would like to tutor you in complex domain and functions if you need help.
I do this problem this way
z=[(-7)3*(cos(π/2+2πk)+i*sin(π/2+2πk))](1/3) k=0,1,2 or
z1=-7(cos(π/6)+i*sin(π/6)) =-7(√3/2+i*1/2)
z2=-7(cos(5π/6)+i*sin(5π/6) =-7(-√3/2+i*1/2)`
z3=-7(cos(9π6)+i*sin(9π/6)= -7*-i=7*i
Dayv O.
sorry again with comma okay --- k=0,1.2 so k takes on those values02/27/21
Dayv O.
meant k= 0,102/27/21