Question

solve the equation z^7=1​

Answer 1

Answer:

z=7√1

seventh root of 1 is one

so

z=1

Answer 2

One solution was found :

z = 1
Step by step solution :

Step 1 :

Polynomial Roots Calculator :

1.1 Find roots (zeroes) of : F(z) = z7-1
Polynomial Roots Calculator is a set of methods aimed at finding values of z for which F(z)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers z which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 1 and the Trailing Constant is -1.

The factor(s) are:

of the Leading Coefficient : 1
of the Trailing Constant : 1

Let us test ....

P Q P/Q F(P/Q) Divisor
-1 1 -1.00 -2.00
1 1 1.00 0.00 z-1

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
z7-1
can be divided with z-1

Polynomial Long Division :

1.2 Polynomial Long Division
Dividing : z7-1
("Dividend")
By : z-1 ("Divisor")

dividend z7 - 1
- divisor * z6 z7 - z6
remainder z6 - 1
- divisor * z5 z6 - z5
remainder z5 - 1
- divisor * z4 z5 - z4
remainder z4 - 1
- divisor * z3 z4 - z3
remainder z3 - 1
- divisor * z2 z3 - z2
remainder z2 - 1
- divisor * z1 z2 - z
remainder z - 1
- divisor * z0 z - 1
remainder 0
Quotient : z6+z5+z4+z3+z2+z+1 Remainder: 0

Polynomial Roots Calculator :

1.3 Find roots (zeroes) of : F(z) = z6+z5+z4+z3+z2+z+1

See theory in step 1.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 1.

The factor(s) are:

of the Leading Coefficient : 1
of the Trailing Constant : 1

Let us test ....

P Q P/Q F(P/Q) Divisor
-1 1 -1.00 1.00
1 1 1.00 7.00

Polynomial Roots Calculator found no rational roots

Equation at the end of step 1 :

(z6 + z5 + z4 + z3 + z2 + z + 1) • (z - 1) = 0
Step 2 :

Theory - Roots of a product :

2.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Equations of order 5 or higher :

2.2 Solve z6+z5+z4+z3+z2+z+1 = 0

In search of an interavl at which the above polynomial changes sign, from negative to positive or the other wayaround.

Method of search: Calculate polynomial values for all integer points between z=-20 and z=+20

No interval at which a change of sign occures has been found. Consequently, Bisection Approximation can not be used. As this is a polynomial of an even degree it may not even have any real (as opposed to imaginary) roots

Solving a Single Variable Equation :

2.3 Solve : z-1 = 0

Add 1 to both sides of the equation :
z = 1

One solution was found :

z = 1

Processing ends successfully