I have been solving the cubic polynomial equation. The simplest of this is
$x^3-1=0$
On solving this equation considering only the real values, I got only one solution i.e
$x=1$
But a cubic polynomial equation has atleast 3 roots(may be complex). I came to realise that the other roots are
$x=\frac{-1±i\sqrt{3}}{2}$
What the point is I got this from considering the x values as complex numbers.
When I try to rewrite the equation in terms of complex variable, this turns into
$z^3-1=0\\z=x+iy\\(x+iy)^3-1=0\\x^3-3xy^2-1 + i(3x^2y-y^3)=0\\$
On equating this to RHS.
REAL PART
$x^3-3xy^2=1$
IMAGINARY PART
$3x^2y-y^3=0$
On solving this imaginary part I get three solutions.
$y=0,±\sqrt{3}x$
When I substitute these into the real part equation.
I get the another equation as follows.
First one $y=0$
The equation is $x^3=1$
This has solution 1
Second and third are $±\sqrt{3}x$
$x^3-9x^3-1=0\\-8x^3=1\\x=\frac{-1}{2}$
So the roots are $1,\frac{-1±i\sqrt{3}}{2}$
Is this a coincidence or any theorem or method.please anyone can explain me what's happened.
Here is what happened when I plot the functions in 3D plotter. Here is plotted function of real part and imaginary part
When I looked the 3D plot from the top(only the curves of the surfaces when the xy plane cut the surface at $z=0$), this is what I saw
The intersection points of the surfaces (imaginary part and real part) and the xy plane are the roots of the equation $z^3-1=0$