Solveeit Logo
Login

Question

Question: Number of values of x (real or complex) simultaneously satisfying the system of equations 1 + ...

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

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

Explanation

Solution

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+z2+z3+........+z17=0...........(1)1 + z + {z^2} + {z^3} + ........ + {z^{17}} = 0...........(1)
1+z+z2+z3+........+z13=0..........(2)1 + z + {z^2} + {z^3} + ........ + {z^{13}} = 0..........(2)
From (2), it can be seen that the sum of terms till z13{z^{13}} is 0.
Remaining terms from (1) are:
z14+z15+z16+z17=0{z^{14}} + {z^{15}} + {z^{16}} + {z^{17}} = 0
Taking z14{z^{14}} common, we get:
z14(1+z+z2+z3)=0{z^{14}}(1 + z + {z^2} + {z^3}) = 0
Factorising:
z14[(1+z2)+z(1+z2)]=0{z^{14}}\left[ {(1 + {z^2}) + z(1 + {z^2})} \right] = 0
z14[(1+z2)(1+z)]=0{z^{14}}\left[ {(1 + {z^2})(1 + z)} \right] = 0
z14[z2−(−1)2(1+z)]=0{z^{14}}\left[ {\\{ {z^2} - {{(\sqrt { - 1} )}^2}\\} (1 + z)} \right] = 0
z14[(z+i)(z−i)(1+z)]=0{z^{14}}\left[ {(z + i)(z - i)(1 + z)} \right] = 0
(As −1=i and using a2−b2=(a+b)(a−b)]({\text{As }}\sqrt { - 1} = i{\text{ and using }}{{\text{a}}^2} - {b^2} = (a + b)(a - b)]
From here, we get:
z14=0,z+i=0,z−i=0,z+1=0{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+z2+.........z13=01 + z + {z^2} + .........{z^{13}} = 0
1+0+0......0≠01 + 0 + 0......0 \ne 0
Does not satisfy the equation.

z=−i⇒z2=i2=−1z = - i \Rightarrow {z^2} = {i^2} = - 1
1+z+z2+.........z13=01 + z + {z^2} + .........{z^{13}} = 0
1−i−1+i+z12+z131 - i - 1 + i + {z^{12}} + {z^{13}}
All terms cancel out except z12 and z13.
z12+z13=0{z^{12}} + {z^{13}} = 0
1−i≠01 - i \ne 0
Does not satisfy the equation.

z=i⇒z2=i2=−1z = i \Rightarrow {z^2} = {i^2} = - 1
1+z+z2+.........z13=01 + z + {z^2} + .........{z^{13}} = 0
All terms cancel out except z12 and z13.
z12+z13=0{z^{12}} + {z^{13}} = 0
1+i≠01 + i \ne 0
Does not satisfy the equation.

z=−1z = - 1
1+z+z2+.........z13=01 + z + {z^2} + .........{z^{13}} = 0
1−1+1−1+1......−1=01 - 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 −1\sqrt { - 1} .
It's important values to remember are:
i2=−1,i4=1{i^2} = - 1,{i^4} = 1
And all multiples of 4 in its power give the answer 1 and multiples of 2 give (-1).