If α is an unreal constant such that αz2+z+¯¯¯¯α=0 has a real root, then
Let a=x+iya=x+iy where x,y∈Rx,y∈R
Then the equation becomes
0+0i=0=(x+iy)z2+z+(x−iy)
=(xz2+z+x)+i(yz2−y)
0+0i=0=(x+iy)z2+z+(x−iy)
=(xz2+z+x)+i(yz2−y)
Since (xz2+z+x)and (yz2−y)are real, we have a real solution to our equation if the equations
xz2+z+x=0 and y(z+1)(z−1) have a common real solution.
If y=0y=0 and x=0x=0, we have z=0z=0 as a real solution. In this case 2x=02x=0
Otherwise if y=0y=0 and x≠0x≠0,
Every real number zz satisfies y(z+1)(z−1)=0y(z+1)(z−1)=0
Thus, the original equation has a real solution iff the equation
xz2+z+x=0 has a real solution.
0=xz2+z+x=x(z+12x)2−14x+x
Or (z+12x)2=14x2−1
xz2+z+x=0
0=xz2+z+x=x(z+12x)2−14x+x
(z+12x)2=14x2−1
Thus, we have a real solution iff
$\dfrac{1}{4x^{2}}-1 >= 0$
or, x2 ≤ (12)2
or, xϵ[−12,12]-{0}
Therefore, 2xϵ[−1,1]-{0}
14x2−1≥0
x2≤(12)2
x∈[−12,12]−{0}
2x∈[−1,1]−{0}
If y≠0y≠0
z=1z=1 or z=−1z=−1
If z=1z=1 is a solution, then x+1+x=0x+1+x=0, or 2x=−12x=−1
If z=−1z=−1 is a solution, then x−1+x=0x−1+x=0, or 2x=12x=1
Therefore, the original equation has a real solution iff
a+a¯=2Re(a)=2x∈[−1,1]