✔ 最佳答案
-z
=-r(cosθ-isinθ)
=r[cos(θ-π)+i(sin(θ-π)]
=r[cos(θ-π+2kπ)+i(sin(θ-π+2kπ)]
where k is an integer such that
0<=θ-π+2kπ<2π
Thus
Arg(-z)=θ-π+2kπ=Argz-π+2kπ
(b)
if |z|=|w|=1
let
z=cosθ1+isinθ1
w=cosθ2+isinθ2
z+w=(cosθ1+cosθ2)+i(sinθ1+sinθ2)
tan[Arg(z+w)]
=(sinθ1+sinθ2)/(cosθ1+cosθ2)
={2[sin(θ1+θ2)/2][cos(θ1-θ2)/2]}/{2[cos(θ1+θ2)/2][cos(θ1-θ2)/2]}
=tan(θ1+θ2)/2
so
Arg(z+w)=(θ1+θ2)/2+kπ
for some integer k such that
0<=(θ1+θ2)/2+kπ<π
Hence
2Arg(z+w)=Argz+Argw+2kπ
for some integer k
(c)
let
z=r1(cosθ1+isinθ1)
w=r2(cosθ2+isinθ2)
|z+w|^2
=[r1cosθ1+r2cosθ2]^2+[r1sinθ1+r2sinθ2]^2
=r1^2+2r1r2(cosθ1cosθ2+sinθ1sinθ2)+r2^2
=r1^2+2r1r2(cos(θ1-θ2))+r2^2
<=r1^2+2r1r2+r2^2
=(r1+r2)^2
=(|z|+|w|)^2
SO
|z+w|^2<=(|z|+|w|)^2
