Math Is Fun Forum

I was right from the begining!
If x=Root[a == b x^2 + c x^3,1](means the first root of f(x)) then Im[x]->0!

Simplifyed:

I'll try something...

I'm sorry. Actually this runction has three roots.
1-real and 2-complex.
Here's it's roots:

Symplifyng...

If we use polar coordinates is easy:
Let A= (r, alfa)pol
Then
A'= (r,alfa+45 deg)pol= (r,alfa+(1/8)2Pi rad)pol= (r,alfa+Pi/4)pol.
Now from decart c.s. to polar c.s.:
Let A= (x,y)dec
r=sqrt(x^2+y^2)
cos(alfa)=y/r
alfa=arccos(y/r)=arccos(y/sqrt(x^2+y^2))

Then
A'= (sqrt(x^2+y^2),arccos(y/sqrt(x^2+y^2))+Pi/4)pol
Now from polar to decart:

x'=r'*sin(alfa')=sqrt(x^2+y^2)*sin(arccos(y/sqrt(x^2+y^2))+Pi/4)
y'=r'*cos(alfa')=sqrt(x^2+y^2)*cos(arccos(y/sqrt(x^2+y^2))+Pi/4)

It's just coordinate change.

Is this write?
I just orbitained it:

So the vector multiplicated by i becomes a vector with negative length!

You have to get all squares with 1.

2) It is not monomial It cannot be simplified:

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