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z^3=1

z^3=1 equation

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Numerical solution:

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The solution

You have entered [src] 
 3    
z  = 1
z3=1z^{3} = 1
Simplify
Detail solution
Given the equation
z3=1z^{3} = 1
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
z33=13\sqrt[3]{z^{3}} = \sqrt[3]{1}
or
z=1z = 1
We get the answer: z = 1

All other 2 root(s) is the complex numbers.
do replacement:
w=zw = z
then the equation will be the:
w3=1w^{3} = 1
Any complex number can presented so:
w=reipw = r e^{i p}
substitute to the equation
r3e3ip=1r^{3} e^{3 i p} = 1
where
r=1r = 1
- the magnitude of the complex number
Substitute r:
e3ip=1e^{3 i p} = 1
Using Euler’s formula, we find roots for p
isin⁡(3p)+cos⁡(3p)=1i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = 1
so
cos⁡(3p)=1\cos{\left(3 p \right)} = 1
and
sin⁡(3p)=0\sin{\left(3 p \right)} = 0
then
p=2πN3p = \frac{2 \pi N}{3}
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
w1=1w_{1} = 1
w2=−12−3i2w_{2} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}
w3=−12+3i2w_{3} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}
do backward replacement
w=zw = z
z=wz = w

The final answer:
z1=1z_{1} = 1
z2=−12−3i2z_{2} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}
z3=−12+3i2z_{3} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}
Vieta's Theorem
it is reduced cubic equation
pz2+qz+v+z3=0p z^{2} + q z + v + z^{3} = 0
where
p=bap = \frac{b}{a}
p=0p = 0
q=caq = \frac{c}{a}
q=0q = 0
v=dav = \frac{d}{a}
v=−1v = -1
Vieta Formulas
z1+z2+z3=−pz_{1} + z_{2} + z_{3} = - p
z1z2+z1z3+z2z3=qz_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q
z1z2z3=vz_{1} z_{2} z_{3} = v
z1+z2+z3=0z_{1} + z_{2} + z_{3} = 0
z1z2+z1z3+z2z3=0z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0
z1z2z3=−1z_{1} z_{2} z_{3} = -1
The graph
-12.5-10.0-7.5-5.0-2.50.02.55.07.510.012.515.0-25002500
Plot the graph f1(x)
Plot the graph f2(x)
Sum and product of roots [src] 
sum
              ___             ___
      1   I*\/ 3      1   I*\/ 3 
1 + - - - ------- + - - + -------
      2      2        2      2
(1+(−12−3i2))+(−12+3i2)\left(1 + \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right)\right) + \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) 
=
0
00 
product
/          ___\ /          ___\
|  1   I*\/ 3 | |  1   I*\/ 3 |
|- - - -------|*|- - + -------|
\  2      2   / \  2      2   /
(−12−3i2)(−12+3i2)\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) 
=
1
11 
1
Rapid solution [src] 
z1 = 1
z1=1z_{1} = 1 
               ___
       1   I*\/ 3 
z2 = - - - -------
       2      2
z2=−12−3i2z_{2} = - \frac{1}{2} - \frac{\sqrt{3} i}{2} 
               ___
       1   I*\/ 3 
z3 = - - + -------
       2      2
z3=−12+3i2z_{3} = - \frac{1}{2} + \frac{\sqrt{3} i}{2} 
z3 = -1/2 + sqrt(3)*i/2
Numerical answer [src] 
z1 = 1.0
z2 = -0.5 - 0.866025403784439*i
z3 = -0.5 + 0.866025403784439*i
z3 = -0.5 + 0.866025403784439*i
The graph
z^3=1 equation
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