Question

Number of values of x (real or complex) simultaneously satisfying the system of equations

 1 + z + {z^2} + {z^3} + ........ + {z^{17}} = 0 $And$ 1 + z + {z^2} + {z^3} + ........ + {z^{13}} = 0 $
A.-1
B.2
C.3
D.4

Answer 1

Hint: We will use the given equations to find the value of 1 through simplification. These values will then be distributed in any equation to check whether it satisfies the system or not.

Complete step-by-step answer:
We have,
 $1+z+z^2+z^3+........+z^{17}=0...........(1)$
 $1+z+z^2+z^3+........+z^{13}=0..........(2)$
From (2), it can be seen that the sum of terms till $z^{13}$ is 0.
Remaining terms from (1) are:
 $z^{14}+z^{15}+z^{16}+z^{17}=0$
Taking $z^{14}$ common, we get:
 $z^{14}(1+z+z^2+z^3)=0$
Factorising:
 $z^{14}\left[(1+z^2)+z(1+z^2)\right]=0$
 $z^{14}\left[(1+z^2)(1+z)\right]=0$
 $z^{14}\left[\{z^2-{(\sqrt{-1})}^2\}(1+z)\right]=0$
 $z^{14}\left[(z+i)(z-i)(1+z)\right]=0$
 $(\text{As }\sqrt{-1}=i\text{ and using }{\text{a}}^2-b^2=(a+b)(a-b)]$
From here, we get:
 $z^{14}=0,z+i=0,z-i=0,z+1=0$
Values of z obtained are:
z = 0,-i, i , -1
Substituting these in (2) and checking which will satisfy the system of equations.
z = 0:
 $1+z+z^2+.........z^{13}=0$
 $1+0+0......0\ne 0$
Does not satisfy the equation.

 $z=-i\Rightarrow z^2=i^2=-1$
 $1+z+z^2+.........z^{13}=0$
 $1-i-1+i+z^{12}+z^{13}$
All terms cancel out except z12 and z13.
$$z^{12}+z^{13}=0$$
 $1-i\ne 0$
Does not satisfy the equation.

 $z=i\Rightarrow z^2=i^2=-1$
 $1+z+z^2+.........z^{13}=0$
All terms cancel out except z12 and z13.
$$z^{12}+z^{13}=0$$
 $1+i\ne 0$
Does not satisfy the equation.

 $z=-1$
 $1+z+z^2+.........z^{13}=0$
 $1-1+1-1+1......-1=0$
Satisfy the equation.
Therefore out of all the values of z, only ‘-1’ satisfies the system of equations and hence the correct option is A)-1
So, the correct answer is “Option A”.

Note: ‘i’ used here stands for iota and is used for representing $\sqrt{-1}$ .
It's important values to remember are:
 $i^2=-1,i^4=1$
And all multiples of 4 in its power give the answer 1 and multiples of 2 give (-1).