# 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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## 3 Answers

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

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

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

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Can you please clarify what you mean my complex conjugate of the entire equation? Complex conjugate of a complex number, x+yi as I understand it is x-yi. How does this translate to an equation? – mathguy80 May 27 '11 at 12:22
I've edited my answer to be more clear. The expression $az^2 + bz + c$ is a complex number, as is $0$. So you can take their complex conjugates. You're doing the same thing to both sides of the equation, so you get another valid equation. You know that $0^*=0$, so $(az^2+bz+c)^*=0$. Now you just need to expand out that bracket using the rules in my answer. – Chris Taylor May 27 '11 at 12:42
Much clearer now. Thank you. I was able to work it out with these 2 identities and @lhf reply. – mathguy80 May 27 '11 at 13:59

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.

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Thanks, you made the factoring look simple. – mathguy80 May 27 '11 at 14:02

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 $y_1=0$.
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I am not sure how this translates to my question. Can you please clarify? Thanks. – mathguy80 May 27 '11 at 12:23
@Américo That's true when $z^* \ne z$. Don't forget the other case. – Bill Dubuque May 27 '11 at 12:26
If $az_1^2 + bz_1 + c = 0$, then $az_2^2 + bz_2 + c = 0$, where $z_2=z_1^{\ast}$ and vice-versa. I used the variables $z_1,z_2$, instead of $z,z^{\ast}$. – Américo Tavares May 27 '11 at 12:29
@Américo The other root need not be $z^*$ when $z^* = z$. Also, don't forget the case $a = 0$. – Bill Dubuque May 27 '11 at 12:32
@Bill: If $z^{\ast}=z$, then the quadratic equation has a double real root, which means that $y=0$, isn't it? – Américo Tavares May 27 '11 at 12:38