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).