Complex numbers equation!

Question

anonymous · Answer

$$((x+iy)-(x-iy))(z ^{4}+2z ^{2}+z)=0$$

anonymous · Answer

One is a normal z=x+iy while the other is a conjugated z=x-iy .
I am referring to the first part of the equation.

anonymous · Answer

so did u try expanding ?

anonymous · Answer

i guess the first part can be written as x² - iy²

kainui · Answer

Just to be clear, this is the same equation right?$$(z-z^*)(z^4+2z^2+z)=0$$

kainui · Answer

first thing to do is to factor out a z from the right part so that you have one of the roots already, z=0. That'll leave you with $$(z-z*)(z^3+2z+1)z=0$$ which is  pretty nice. Really since this is all multiplied out together, each individual part has to be zero on its own to make a root, so, $$z-z^*=0$$$$z^3+2z+1=0$$
So for the first one, when is a complex number equal to its conjugate? When the imaginary part is 0 since bi=-bi when b=0. So essentially all real numbers make this a true statement, conveniently. On to the next equation, maybe we can get some more roots, but complex this time. 

There are more but I'm not really in a rush since no one's here.

anonymous · Answer

Okay,
I tried solving the second equation but I didn't get it so I still back to square one.
Thank you by the way for helping me out.

anonymous · Answer

well covert it to :-
z^5+2z^3+z^2-(z^3|z|+2z|z|+|z|)=0
at least one of the sol
z=0

anonymous · Answer

@BSwan I don't get it >:D Isn't there a simpler passage?