设z=12+32i（i是虚数单位），则z+2z2+3z3+4z4+5z5+6z6=（）A．6zB．6z2C．6.zD．-6z

Question

设z=(1)/(2)+(√(3))/(2)i（i是虚数单位），则z+2z²+3z³+4z⁴+5z⁵+6z⁶=（　　）

A．6z
B．6z²
C．6

.

z

D．-6z

Answer 1

∵z=(1)/(2)+(√(3))/(2)i=cos(π)/(3)+isin(π)/(3)，
z+2z²+3z³+4z⁴+5z⁵+6z⁶=cos(π)/(3)+isin(π)/(3)+2cos(2π)/(3)+2sin(2π)/(3)i+3cosπ
+3sinπi+4cos(4π)/(3)+4sin(4π)/(3)i+5cos(5π)/(3)+5sin(5π)/(3)i+6cos2π+6sin2πi
=6（(1)/(2)-(√(3))/(2)i）=6

.

z

故选C．