设z=12+32i(i是虚数单位),则z+2z2+3z3+4z4+5z5+6z6=( ) A.6z B.6z2 C.6.z D.-6z
问题描述:
设z=
+1 2
i(i是虚数单位),则z+2z2+3z3+4z4+5z5+6z6=( )
3
2
A. 6z
B. 6z2
C. 6
. z
D. -6z
答
∵z=12+32i=cosπ3+isinπ3,z+2z2+3z3+4z4+5z5+6z6=cosπ3+isinπ3+2cos2π3+2sin2π3i+3cosπ+3sinπi+4cos4π3+4sin4π3i+5cos5π3+5sin5π3i+6cos2π+6sin2πi=6(12-32i)=6.z故选C.