Complex Quadratic Formula for Complex Numbers

In summary, the equation az2+ bz+ c= 0 has two complex roots, w_1, w_2, which are the square roots of b^2- 4ac.
  • #1
e(ho0n3
1,357
0
[SOLVED] Complex Quadratic Formula

Homework Statement
Let a, b and c be complex numbers with a not equal to 0. Show that the solution of [itex]az^2 + bz + c = 0[/itex] are [itex]z_1, z_2 = (-b \pm \sqrt(b^2 - 4ac))/(2a)[/itex]

The attempt at a solution
I'm assuming z is also complex. Multiplying the equation by 4ac and completing the square yields [itex](2az + b)^2 = b^2 - 4ac[/itex] There are two complex numbers, [itex]w_1, w_2[/itex], that are the square roots of [itex]b^2 - 4ac[/itex] so setting [itex]2az_j + b = w_j[/itex] and solving for [itex]z_j[/itex] yields

[tex]z_j = \frac{-b + w_j}{2a}[/tex]

for j = 1, 2. The section where I got this problem from does not define the square root of a complex numbers; it only talks about the roots of complex numbers. Would it be correct to stop here and consider the problem solved?
 
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  • #2
How about just putting
[tex]\frac{-b+ \sqrt{b^2- 4ac}}{2a}[/tex]
and
[tex]\frac{-b- \sqrt{b^2- 4ac}}{2a}[/tex]
into the equation az2+ bz+ c= 0 and see what happens?

(I am puzzled about " does not define the square root of a complex numbers; it only talks about the roots of complex numbers". How does it talk about the roots of complex numbers without defining them? In any case, I would think the definition is obvious: a complex number, z, is a square root of the complex number a if and only if z2= a.)
 
  • #3
My confusion stems from how the radical symbol extends to complex numbers. So the notation [itex]\sqrt{z}[/itex] is the principal square root of z? Which of the two roots is the principal though? The one with the same signs in the real and imaginary part as z?
 
  • #4
e(ho0n3 said:
My confusion stems from how the radical symbol extends to complex numbers. So the notation [itex]\sqrt{z}[/itex] is the principal square root of z? Which of the two roots is the principal though? The one with the same signs in the real and imaginary part as z?

[itex]\sqrt{z}[/itex]

z is a complex number of the form a+bi


[itex]\sqrt{z}[/itex] is different from a complex number,z, being a root of a quadratic.
 
  • #5
rock.freak667 said:
[itex]\sqrt{z}[/itex] is different from a complex number,z, being a root of a quadratic.

Yes, I know it is different. That doesn't answer my question.
 
  • #6
Pick one - the positive one, or the one with a positive imaginary part.

The roots will still be [itex]\pm \sqrt{z}[/itex].
 
  • #7
Just as for real numbers, every complex number has two square roots. If [itex]z=re^{i\theta}[/itex], with [itex]r[/itex] real and positive and [itex]-\pi\le\theta<\pi[/itex], the two square roots are [itex]+\sqrt{r}e^{i\theta/2}[/itex] and
[itex]-\sqrt{r}e^{i\theta/2}[/itex]. The first of these is usually taken to be the principal root. There is a "branch cut" along the negative real axis; the square root of z changes sign abruptly as z moves across the negative real axis. Although this is the conventional choice, the branch cut could instead be chosen to lie along any line (straight or curved) from 0 to infinity.
 
  • #8
e(ho0n3 said:
My confusion stems from how the radical symbol extends to complex numbers. So the notation [itex]\sqrt{z}[/itex] is the principal square root of z? Which of the two roots is the principal though? The one with the same signs in the real and imaginary part as z?

Wikipedia says put the branch cut along the negative real axis, so define arg(z) to be between -pi and pi and then define sqrt(z) to be the root with argument arg(z)/2. But that's not the only choice. I think it's better to define what you mean rather than assuming that 'everyone knows'.
 
  • #9
Since the complex numbers are not an "ordered" field there is no "principle" root except, as Dick says, by some arbitrary choice- that in general cannot be extended to other roots.

And, it is not necessary for the problem you stated. Your problem involves both roots and it doesn't matter which you consider "principle". The only thing you need to know about [itex]\sqrt{b^2- 4ac}[/itex] is that [itex]\left(\sqrt{b^2- 4ac}\right)^2= b^2- 4ac[/itex].
 
  • #10
OK. I think I'm satisfied with the explanations. Thank you everyone.
 

What is the Complex Quadratic Formula?

The Complex Quadratic Formula is a mathematical equation used to find the roots or solutions of a quadratic equation with complex numbers. It is an extension of the standard quadratic formula, which only applies to equations with real number solutions.

What is the general form of the Complex Quadratic Formula?

The general form of the Complex Quadratic Formula is:
x = (-b ± √(b²-4ac)) / 2a
where a, b, and c are coefficients of the quadratic equation in the form of ax² + bx + c = 0.

How is the Complex Quadratic Formula derived?

The Complex Quadratic Formula is derived from the standard quadratic formula by extending the concept of square roots to include complex numbers. This is done by using the quadratic formula to solve for the roots of the equation (x² +1 = 0) and then generalizing the resulting formula to apply to all quadratic equations with complex solutions.

Can the Complex Quadratic Formula be used to find both real and complex solutions?

Yes, the Complex Quadratic Formula can be used to find both real and complex solutions. If the discriminant (b²-4ac) is negative, the formula will result in two complex solutions. If the discriminant is zero, the formula will result in one real solution. And if the discriminant is positive, the formula will result in two real solutions.

What are some practical applications of the Complex Quadratic Formula?

The Complex Quadratic Formula has many practical applications in fields such as physics, engineering, and economics. It can be used to solve problems involving complex numbers, such as calculating electrical impedance or predicting stock market trends. It is also used in computer graphics to create 3D animations and in cryptography to encrypt and decrypt data.

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