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z^3=i equation

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Numerical solution:

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The solution

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 3    
z  = I
z3=iz^{3} = i
Simplify
Detail solution
Given the equation
z3=iz^{3} = i
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=i3\sqrt[3]{z^{3}} = \sqrt[3]{i}
or
z=i3z = \sqrt[3]{i}
Expand brackets in the right part
z = i^1/3

We get the answer: z = i^(1/3)

All other 3 root(s) is the complex numbers.
do replacement:
w=zw = z
then the equation will be the:
w3=iw^{3} = i
Any complex number can presented so:
w=reipw = r e^{i p}
substitute to the equation
r3e3ip=ir^{3} e^{3 i p} = i
where
r=1r = 1
- the magnitude of the complex number
Substitute r:
e3ip=ie^{3 i p} = i
Using Euler’s formula, we find roots for p
isin⁡(3p)+cos⁡(3p)=ii \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = i
so
cos⁡(3p)=0\cos{\left(3 p \right)} = 0
and
sin⁡(3p)=1\sin{\left(3 p \right)} = 1
then
p=2πN3+π6p = \frac{2 \pi N}{3} + \frac{\pi}{6}
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=−iw_{1} = - i
w2=−32+i2w_{2} = - \frac{\sqrt{3}}{2} + \frac{i}{2}
w3=32+i2w_{3} = \frac{\sqrt{3}}{2} + \frac{i}{2}
do backward replacement
w=zw = z
z=wz = w

The final answer:
z1=−iz_{1} = - i
z2=−32+i2z_{2} = - \frac{\sqrt{3}}{2} + \frac{i}{2}
z3=32+i2z_{3} = \frac{\sqrt{3}}{2} + \frac{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=−iv = - i
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=−iz_{1} z_{2} z_{3} = - i
The graph
Sum and product of roots [src] 
sum
           ___         ___
     I   \/ 3    I   \/ 3 
-I + - - ----- + - + -----
     2     2     2     2
(−i+(−32+i2))+(32+i2)\left(- i + \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right)\right) + \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) 
=
0
00 
product
   /      ___\ /      ___\
   |I   \/ 3 | |I   \/ 3 |
-I*|- - -----|*|- + -----|
   \2     2  / \2     2  /
−i(−32+i2)(32+i2)- i \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) 
=
I
ii 
i
Rapid solution [src] 
z1 = -I
z1=−iz_{1} = - i 
           ___
     I   \/ 3 
z2 = - - -----
     2     2
z2=−32+i2z_{2} = - \frac{\sqrt{3}}{2} + \frac{i}{2} 
           ___
     I   \/ 3 
z3 = - + -----
     2     2
z3=32+i2z_{3} = \frac{\sqrt{3}}{2} + \frac{i}{2} 
z3 = sqrt(3)/2 + i/2
Numerical answer [src] 
z1 = -0.866025403784439 + 0.5*i
z2 = 0.866025403784439 + 0.5*i
z3 = -1.0*i
z3 = -1.0*i
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