已知复数z=(√3+i)/((1-√3i)^2 ),则|z|=(A) 1/4 (B) 1/2 (C)1 (D)2

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已知复数z=(√3+i)/((1-√3i)^2),则|z|=(A)1/4(B)1/2(C)1(D)2已知复数z=(√3+i)/((1-√3i)^2),则|z|=(A)1/4(B)1/2(C)1(D)2

已知复数z=(√3+i)/((1-√3i)^2 ),则|z|=(A) 1/4 (B) 1/2 (C)1 (D)2
已知复数z=(√3+i)/((1-√3i)^2 ),则|z|=
(A) 1/4 (B) 1/2 (C)1 (D)2

已知复数z=(√3+i)/((1-√3i)^2 ),则|z|=(A) 1/4 (B) 1/2 (C)1 (D)2
z=(√3+i)/[1-√3i)²]
=(√3+i)/(-2-2√3i)
=-(√3+i)/[2(1+√3i)]
=-(√3+i)(1-√3i)/4
=-(2√3-2i)/4
=-(√3-i)/4
|z|=(1/4)√[(√3)²+(-1)²]=2/4=1/2
选B

选B