roots of equation lie on $x^2+y^2=1$

Question

If $a,b\in\mathbb{C}$ and $z$ is a non zero complex number.

The the rots of the equation $az^3+bz^2+\bar{b}z+\bar{a}=0$

lie on $x^2+y^2=1$. Where $z=x+iy$

Plan: I am taking $z=x+iy$

$\displaystyle a(x+iy)^3+b(x+iy)^2+\bar{b}(x+iy)+\bar{a}=0$

$a(x^3-iy^3+3ixy(x+iy))+b(x^2-y^2+2ixy)+\bar{b}(x+iy)+\bar{a}=0$

$\displaystyle a(x^3-3xy^2)+b(x^2-y^2)+\bar{b}x+\bar{a}=0$

and $-ay^3+3x^2y+2bxy+\bar{b}x=0$

how do i solve it help me please

Answer 1

I prove that if $a,b$ are complex numbers roots the equation (1) are pairs of the type: $z$ and $1/\bar{z}$ which means all roots are uni-modular or at least one (un-paired one) is so ($x^2+y^2=1$).

Let $$az^3+bz^2+\bar{b}z+\bar{a}=0.....(1).$$ Complex conjugation of Eq. (1) gives $$\bar{a} \bar{z}^3+\bar{b}{\bar{z}^2}+b \bar{z}+a=0....(2).$$ Now if we change $\bar{z}=\frac{1}{z}$ in Eq. (2), we get $$\frac{\bar{a}}{z^3}+\frac{\bar{b}}{z^2}+\frac{b}{z}+\frac{a}{z}=0...(3).$$ Which (3) is nothing but Eq. (1). This proves that Eq. (1) has all roots sucj that $x^2+y^2=1$. In other words roots of Eq. (1), are pairs of the type $z$ and $1/\bar{z}$, which means either all roots are uni-modular or at least one (the unpaired one) is so.

For instance one may check that for $a=1-i$ and $b=i$,we get roots as $-(1/4 + i/4) (i + \sqrt{7}),~(1/4 + i/4) (-i + \sqrt{7}),~ -1$.

Answer 2

Put $a=1$ and $b=4$. You find roots on the real axis which are not $\pm1$; to wit, $z\in\{-1,(-3\pm\sqrt{5})/2\}$ and only $-1$ meets the claim. You'll have to put constraints on $a$ and $b$ if this is to work.

What always does work is, at least one root will lie on the unit circle. In the above example that would be $-1$.

Answer 3

I now prove the conjecture of Oscar given in his answer. Oscar's conjecture limits the claim of the original proposal by Jacky, and hence it corrects the original question.

The Conjecture of Oscar Lanzi:

If $a,b$ are complex numbers, the equation $$ az^3+bz^2+\bar{b}z+\bar{a}=0.....(1),$$ has at least one root which is uni-modular.

The Proof:

Let $t$ be one root of (1), then $$ at^3+bt^2+\bar{b} t+ \bar{a}=0....(2)$$ Complex conjugation of (2) leads to $$\bar{a}\bar{t}^3+\bar{b}\bar{t}^2+b \bar{t} +a=0....(3).$$ Multiplying Eq. (3) by $t^3$, we get $$\bar{a}(t\bar{t})^3+\bar{b} t({t \bar {t}})^2+ b t^2 (t \bar{t})+a t^3=0...(4). $$ This equation can be re-written as $$ \bar{a} |t|^6+\bar{b} t|t|^4+b t^2 |t|^2+ a t^3=0...(5).$$ Subtracting Eqs. (2) and (5), we get $$\bar{a} (|t|^6-1)+\bar{b} t (|t|^4-1)+b t^2 (|t|^2-1)=0....(6).$$ Remarkably, we get $(|t|^2-1)=0.$ Hence proved that at least one root of Eq. (1) is uni-modular: $(x^2+y^2)=1.$