Complex polynomial: If p(z) is a real fifth degree polynomial with an only real root.

[Caracolillo97]

Firstly, I want to say that I am from Spain, so I will do my best to express myself correctly.

If p(z) is a real fifth degree polynomial with an only real root. If p(0)≠0, the quantity of real solutions of p(z-iz)=0 (where z is the conjugate):


a) at least 1
b) at the most 2
c) it can be 4

I was so pleased someone could help me. Thank you.

 

[Ishuda]

Firstly, I want to say that I am from Spain, so I will do my best to express myself correctly.

If p(z) is a real fifth degree polynomial with an only real root. If p(0)≠0, the quantity of real solutions of p(z-iz)=0 (where z is the conjugate):


a) at least 1
b) at the most 2
c) it can be 4

I was so pleased someone could help me. Thank you.

Please read the http://www.freemathhelp.com/forum/announcement.php?f=32 page. What have you done? Even if you think you are wrong, please show us what you have done so that we can find your mistake (if any) or point you toward a solution.


As a hint, consider
u = Real(z-iz)
v = Imag(z-iz)
If p has only real roots what does this say about v if p(u+iv)=0, i.e. what is the relationship between x and y and where in the complex plane must zeros of p(u+iv) lie? If p(u+iv)=0, what does this then also say about the value of u and the relationship between x and y,