Question

The equation z5 + z4 + z3 + z2 + z + 1 = 0 is satisfied by:

Answer 1

It is well known that zn−1=(z−1)(∑n−1i=0zi). Substituting n=5 gives us z5−1=(z−1)(1+z+z2+z3+z4). Thus, we have the following:

1+z+z2+z3+z4=z5−1z−1 (z≠1)
The right side of the equation is 0 only when the numerator z5−1=0 and z≠1. Thus, we must solve the equation z5=1 when z≠1, which is just the complex fifth roots of unity.

By De Moivre's formula, we know these roots of unity are cos2nπ5+isin2nπ5 for n∈Z. However, if n is 0, then this number equals 1, which can not be a solution. Also, if n<0 or n≥5, then we can take the modulo of n by 5 to find an equivalent solution as adding or subtracting n by 5 simply adds or subtracts the angle by 2π, which does not change the answer. Thus, 1≤n≤4 and the solution set is {cos2nπ5+isin2nπ5:n∈{1,2,3,4}}
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Answer 2

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>> z³( z² + z + 1 ) + ( z² + z + 1 ) = 0
=> ( z³ + 1 )( z² + z + 1 ) = 0

=> ( z + 1 )( z² - z + 1 )( z² + z + 1 ) = 0
=> [ z = -1 ][ z = ω , ω² ] , where 'ω' is the cube-root of unity

Note : ( z² - z + 1 ) never gets a Real No. Soln
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