Question

已知复数z=r(cosθ+isinθ)r,θ∈R (1)分别计算z^2,z^3并由此可归纳出z^n

(2)试着根据归纳结果计算（（根号3）+（i））^7

Answer 1

已知复数
z=r(cosθ+isinθ)
z^2=r^2(cosθ+isinθ)^2
=r^2(cos^2θ-sin^2θ+isin2θ)
=r^2(cos2θ+isin2θ)
z^3=z*z^2=r(cosθ+isinθ)*r^2(cos2θ+isin2θ)
=r^3(cosθcos2θ+isin2θcosθ+isinθcos2θ-sinθsin2θ)
=r^3(cos3θ+isin3θ)
由此可归纳出
z^n=r^n(cosnθ+isinnθ)
(√3+i)^7
=2^7(√3/2+1/2i)^7
=2^7(cosπ/6+Isinπ/6)^7
=2^7(cos7π/6+isin7π/6)
=2^7(-(√3/2-1/2i)
=-2^6(√3+i)

Answer 2

（1）z^2=r^2* ((cosθ)^2+i(sinθ)^2)
z^3=r^3* ((cosθ)^3+i(sinθ)^3)
(2) 归纳出来 z^7=r^7*((cosθ)^7+i(sinθ)^7)
cosθ=根号3
sinθ=1 代入自己算算吧

Answer 3

z^2=r^2(cos2θ+isin2θ),z^3=r^3(cos3θ+isin3θ),z^n=r^n(cosnθ+isinnθ);
（根号3）+（i））^7=2^7[cos（7×30度）+isin（7×30度)]=128(cos210度+isin210度)
=64 [-(根号3）-（i）]

Answer 4

(1)z^2=r^2(cos2θ+isin2θ)
z^3=r^3(cos3θ+isin3θ)
由此可归纳出z^n=r^n(cosnθ+isinnθ)
(2)(sqrt(3)+i)^7
={2[cos(π/6)+isin(π/6)]}^7
=2^7[cos(7π/6)+isin(7π/6)]
=128(-sqrt(3)/2-i/2)
=-64sqrt(3)-64i