How to show $ a,b,c \in \mathbb R , z \in \mathbb C , az^2 + bz + c = 0 \iff a\bar{z}^2 + b\bar{z} + c = 0$?

I am completely stumped by with this problem.

If a, b and c are real numbers and z is a complex number of the form $x + yi$, prove that $$az^2 + bz + c = 0 \iff a(z^*)^2 + b(z^*) + c = 0$$

I tried substituting $z = x + yi$ and for $z^*= x – yi$, but that gives a bigger equation with multiple variables. How should I proceed with this? Thanks for your help.

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Try taking the complex conjugate of each side of the equation, and remembering that

$$(w+z)^* = w^* + z^*$$

$$(wz)^* = w^*z^*$$

Here is a solution along the lines you took:

$a(z^*)^2 + b(z^*) + c = a(x-yi)^2 + b(x-yi) + c$

$\quad=a(x^2-2xyi-y^2) + b(x-yi) + c $

$\quad= (a(x^2-y^2)+bx+c) – (2axy+by)i$

$\quad=(az^2 + bz + c)^*$

Since $w=0$ iff $w^*=0$, you get your result.

Note: This answer is based on the relation between the roots of a quadratic equation with real coefficients, in the case when both are complex numbers.

I used below $z_1,z_2$, instead of $z,z^{\ast}$.


If one of the roots of a quadratic equation is the complex number $z_1=x_1+iy_1$ the other root $z_2=x_2+iy_2$ is the conjugate of $z_1$, i.e $z_2=x_1-iy_1=z_1^{\ast}$, as it can be seen, among other ways, by the resolvent formula

$$az^2 + bz + c = 0 \iff z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad(a \ne 0).$$

Thus, if

$$z_1=\frac{-b+\sqrt{b^2-4ac}}{2a}=x_1+iy_1,$$

then

$$z_2=\frac{-b-\sqrt{b^2-4ac}}{2a}=x_1-iy_1=z_1^{\ast}$$

and vice-versa.

Since $z_1,z_2$ are roots, both satisfy the quadratic equation, by definition.


Added in view of the comments below.

  • If $z_2=z_1$, then $z_2=x_2=x_1$ and $y_1=y_2=0$.
  • If $a=0$, then the equation reduces to $bz_1+c=0$, which has a single root $z_1=x_1$, and $y_1=0$.