PRACTICE PROBLEMS ON FINDING ROOTS OF A COMPLEX NUMBER

1)  z₀ = ³√2(cis (2π/3))

z₁ = ³√2(cis (8π/3))

z₂ = ³√2(cis (10π/3))

2)  z₀  =  ³√2(cis (π/12))

z₁  =  ³√2(cis (9π/12))

z₂  =  ³√2(cis (17π/12))

3)  z₀  =  ³√3(cis (4π/9))

z₁  =  ³√3(cis (10π/9))

z₂  =  ³√2[cis (16π/9))

4)  z₀  =  3(cis (11π/18))

z₁  =  3(cis (23π/18))

z₂  =  3(cos (35π/18))

5)  z₀  =  ⁶√8(cis (3π/12))

z₁  =  ⁶√8(cos (11π/12))

z₂  =  ⁶√8(cos (19π/12))

1)  z₀  =  ⁸√2(cis (π/16))

z₁  =  ⁸√2(cis (9π/16))

z₂  =  ⁸√2(cis (17π/16))

z₃  =  ⁸√2(cis (25π/16))

2)  z₀  =  ¹²√2(cis (π/24))

z₁  =  ⁸√2(cis (7π/24))

z₂  =  ¹²√2(cis (5π/8))

z₃  =  ¹²√2(cis (23π/24))

z₄  =  ¹²√2(cis (31π/24))

z₅  =  ¹²√2(cis (39π/24))

3)  z₀  =  ⁶√8(cis (π/12))

z₁  =  ⁶√8(cis (3π/4))

z₂  =  ⁶√8(cis (17π/12))

4)  z₀  =  ⁸√8(cis (3π/16))

z₁  =  ⁸√8(cis (11π/16))

z₂  =  ⁸√8(cis (19π/16))

z₃  =  ⁸√8(cis (27π/16))

1)  z³  =  -√2/2 + i √2/2

2)  z⁵  =  243i

3)  z³  =  4√2 + i 4√2

4)  z⁵  =  -4 - 4i

5)  z³  =  -8