Let z = 1 + ai be a complex number, a 0, such that z3 is a real number. Then the sum 1 + z + z2 +..... + z11 is equal to :

Question

Let z = 1 + ai be a complex number, a > 0, such that z3 is a real number. Then the sum 1 + z + z2 +..... + z11 is equal to : (A) - 1250 √3 i  (B) 1250 √3 i (C) 1365 √3 i (D) - 1365 √3 i

Tardigrade Admin · Accepted Answer

z=1+a i, a > 0 z3=1-3 a2+(3 a-a3) i is a real number ⇒ 3 a-a3=0 ⇒ a2=3 ⇒ a=√3, a > 0 ⇒ z=1+√3 i =2( cos (π/3)+i sin (π/3)) Now 1+z+z2+ ldots ldots+z11=(1(1-z12)/1-z)=(1-212( cos 4 π+i sin 4 π)/1-(1+i √3)) =(1-212/-i √3)=(4095/i √3)=-1365 √3 i

Q. Let $z = 1 + ai$ be a complex number, $a > 0$ , such that $z^{3}$ is a real number. Then the sum $1 + z + z^{2} + ..... + z^{11}$ is equal to :

$- 12503 i$

$12503 i$

$13653 i$

$- 13653 i$

Q. Let z=1+ai be a complex number, a>0, such that z3 is a real number. Then the sum 1+z+z2+.....+z11 is equal to :

−12503​i

12503​i

13653​i

−13653​i

z=1+ai, a>0 z3=1−3a2+(3a−a3)i is a real number ⇒3a−a3=0 ⇒a2=3 ⇒a=3​, a>0 ⇒z=1+3​i =2(cos3π​+isin3π​) Now 1+z+z2+……+z11=1−z1(1−z12)​=1−(1+i3​)1−212(cos4π+isin4π)​ =−i3​1−212​=i3​4095​=−13653​i

Q. Let $z = 1 + ai$ be a complex number, $a > 0$ , such that $z^{3}$ is a real number. Then the sum $1 + z + z^{2} + ..... + z^{11}$ is equal to :

$- 12503 i$

$12503 i$

$13653 i$

$- 13653 i$