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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.

3 Answers 3

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

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

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

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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?2011-05-27
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    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.2011-05-27
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    Much clearer now. Thank you. I was able to work it out with these 2 identities and @lhf reply.2011-05-27
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    If $a=0$ we have nothing to prove. Otherwise $z_1+z_2=-\dfrac{b}{a}\in\Bbb{R}$ and $z_1z_2=\dfrac{c}{a}\in \Bbb{R}.$ Only complex conjugates have this property.2016-10-25
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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.2011-05-27
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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.2011-05-27
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    @Américo That's true when $z^* \ne z$. Don't forget the other case.2011-05-27
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    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 $z,z^{\ast}$.2011-05-27
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    @Américo The other root need not be $z^*$ when $z^* = z$. Also, don't forget the case $a = 0$.2011-05-27
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    @Bill: If $z^{\ast}=z$, then the quadratic equation has a double real root, which means that $y=0$, isn't it?2011-05-27
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    @Bill: If $a=0$, then the equation is $bz+c=0$, which has a single real root, and again $y=0$.2011-05-27
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    @Ame Yes, but one needs to explicitly mention that (easy) case, unless "complex" above means non-real complex. Also one needs to explicitly treat the degenerate case $a = 0$. One can avoid this case analysis by noting that conjugation is an automorphism.2011-05-27
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    @Bill: Thanks! I added these considerations to my answer. I hope it is now complete.2011-05-27
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    @Ame if $z^* = z$ then the equivalence is clear, no need to mention anything about double roots. I just wanted to point out that one does need to mention these (easy) cases for this type of argument to be complete.2011-05-27
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    @Bill: I simplified the answer once more, in view of your comment.2011-05-27
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    Thanks guys. I feel I understood most of what you guys are saying. But need some time to digest it better. :)2011-05-27
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    @mathguy80: my answer is based on the relation between the roots of a quadratic equation with real coefficients, in the case when both are complex numbers.2011-05-27
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    Converted to CW because the reasoning was not clearly worded, though the idea were correct, I think.2011-05-27