rcisθ

解xⁿ=p

當θ=2wπ (1≦w≦n，其中w為正整數)，[cosθ+isinθ]=1

xⁿ=p[cos2wπ+isin2wπ]

∵[cosθ+isinθ]^k=[cos(kθ)+isin(kθ)]，看證明

∴x=p^(1/n)[cos2wπ+isin2wπ]^(1/n)

∴x=p^(1/n)[cos(2wπ/n)+isin(2wπ/n)]

以x⁸=p為例

x=p^(1/8)[cos(2π/8)+isin(2π/8)],p^(1/8)[cos(4π/8)+isin(4π/8)],p^(1/8)[cos(6π/8)+isin(6π/8)],p^(1/8)[cos(8π/8)+isin(8π/8)],

p^(1/8)[cos(10π/8)+isin(10π/8)],p^(1/8)[cos(12π/8)+isin(12π/8)],p^(1/8)[cos(14π/8)+isin(14π/8)],p^(1/8)[cos(16π/8)+isin(16π/8)]

x=p^(1/8)[cos(π/4)+isin(π/4)],p^(1/8)[cos(π/2)+isin(π/2)],p^(1/8)[cos(3π/4)+isin(3π/4)],p^(1/8)[cosπ+isinπ],

p^(1/8)[cos(5π/4)+isin(5π/4)],p^(1/8)[cos(3π/2)+isin(3π/2)],p^(1/8)[cos(7π/4)+isin(7π/4)],p^(1/8)[cos(2π)+isin(2π)]

x=p^(1/8)[2^(1/2)/2+(i2^(1/2))/2],p^(1/8)[i],p^(1/8)[-2^(1/2)/2+i2^(1/2)/2],p^(1/8)[-1],

p^(1/8)[-2^(1/2)/2-i2^(1/2)/2],p^(1/8)[-i],p^(1/8)[2^(1/2)/2-i2^(1/2)/2],p^(1/8)[1]

x=p^(1/8)[2^(1/2)/2+(i2^(1/2))/2],ip^(1/8),p^(1/8)[-2^(1/2)/2+i2^(1/2)/2],-p^(1/8),

p^(1/8)[-2^(1/2)/2-i2^(1/2)/2],-ip^(1/8),p^(1/8)[2^(1/2)/2-i2^(1/2)/2],p^(1/8)