Let z1 and z2 be roots of the equation z² + pz + q = 0, where the coefficients p and q may be comple

Q: Why can't we just use the quadratic formula to find $z_1$ and $z_2$?

The quadratic formula involves square roots of complex numbers, which is algebraically messy and often leads to ambiguity. By using the geometric properties of the roots, we bypass the need to solve for $z_1$ and $z_2$ individually, leading to a much more elegant and direct proof.

Q: Why is $z_2 = z_1 e^{i\alpha}$?

In the complex plane, any complex number $z$ can be written as $re^{i\theta}$. If we rotate this vector by an angle $\alpha$ without changing its magnitude $r$, the new number is $re^{i(\theta + \alpha)} = re^{i\theta} e^{i\alpha} = z e^{i\alpha}$. Since $OA = OB$, the magnitudes are equal, making this rotation perfectly valid.

Q: How did we know to factor out $e^{i\alpha/2}$?

This is a classic 'trick' in complex trigonometry. We want to transform the sum $(1 + e^{i\alpha})$ into a form that matches the cosine function. Factoring out the half-angle $e^{i\alpha/2}$ leaves us with $(e^{-i\alpha/2} + e^{i\alpha/2})$, which is exactly $2\cos(\alpha/2)$ via Euler's formula. It is the key to unlocking the symmetry of the expression.

Question

Let $z_1$ and $z_2$ be roots of the equation $z^2 + pz + q = 0$, where the coefficients $p$ and $q$ may be complex numbers. Let $A$ and $B$ represent $z_1$ and $z_2$ in the complex plane. If $\angle AOB = \alpha 
eq 0$ and $OA = OB$, where $O$ is the origin, prove that $p^2 = 4q \cos^2(\alpha/2)$.

Sigma Percentile · Accepted Answer

### The Geometry of Roots: A Journey into Complex Symmetry

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a quadratic equation; we are uncovering a hidden geometric truth.

We are looking at the equation $z^2 + pz + q = 0$, where $z_1$ and $z_2$ are the roots. Instead of treating them as abstract symbols, let us visualize them as points $A$ and $B$ in the complex plane.

We are told that $OA = OB$ and the angle $\angle AOB = \alpha$. This is our canvas. Let us paint the solution.

### Phase 1: The Geometric Bridge

Imagine standing at the origin $O$. You have two vectors, $z_1$ and $z_2$, stretching out to points $A$ and $B$.

Because the problem states $OA = OB$, we know these vectors have the same magnitude. Because the angle between them is $\alpha$, we can describe $z_2$ as a rotation of $z_1$.

In the complex plane, rotation is elegant—it is simply multiplication by $e^{i\alpha}$. Thus, we establish our first vital relationship:
$$z_2 = z_1 e^{i\alpha}$$

### Phase 2: The Algebraic Foundation

Now, we turn to the algebra. Vieta's formulas are our most trusted tools here. For the quadratic $z^2 + pz + q = 0$, we know two things: the sum of the roots is $-p$, and the product is $q$.

So, we have:
1. $z_1 + z_2 = -p$
2. $z_1 z_2 = q$

Our mission is to connect these to the angle $\alpha$. We have the sum, and we have the rotation relation. Let us combine them.

### Phase 3: The Half-Angle Transformation

Substitute our rotation relation into the sum equation: $-p = z_1 + z_1 e^{i\alpha}$. Factoring out $z_1$, we get $-p = z_1(1 + e^{i\alpha})$.

This looks promising, but how do we get to $\cos(\alpha/2)$? This is where the 'half-angle trick' comes in. We factor out $e^{i\alpha/2}$ from the parenthesis:

$$-p = z_1 e^{i\alpha/2} (e^{-i\alpha/2} + e^{i\alpha/2})$$

Look at that term in the parentheses! Euler's identity tells us that $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$. By setting $	heta = \alpha/2$, the expression simplifies beautifully to $2\cos(\alpha/2)$.

Our equation now reads:
$$-p = z_1 e^{i\alpha/2} (2\cos(\alpha/2))$$

### Phase 4: The Final Synthesis

We are almost there. The proof requires $p^2$. So, let us square both sides of our equation:

$$p^2 = (-2 z_1 e^{i\alpha/2} \cos(\alpha/2))^2$$

$$p^2 = 4 z_1^2 e^{i\alpha} \cos^2(\alpha/2)$$

Now, look closely at the term $z_1^2 e^{i\alpha}$. We can rewrite this as $z_1 (z_1 e^{i\alpha})$.

Recall our rotation relation from Phase 1: $z_2 = z_1 e^{i\alpha}$. Therefore, $z_1^2 e^{i\alpha}$ is simply $z_1 z_2$. And what is $z_1 z_2$? It is $q$, the product of the roots!

Substituting $q$ back into our equation, we arrive at the destination:
**$$p^2 = 4q \cos^2(\alpha/2)$$**

### Conclusion

We started with a simple quadratic and ended with a beautiful trigonometric identity. This is the power of complex numbers—they allow us to weave geometry and algebra into a single, cohesive narrative.

Never fear the complexity of the variables; look for the symmetry, trust the identities, and the path will reveal itself.

Visualized Solution

Visualizing the Roots $z_{1}$ and $z_{2}$

Applying Vieta's Formulas: Sum of Roots

Applying Vieta's Formulas: Product of Roots

Rotation in the Complex Plane

Substituting Rotation into the Sum

The Half-Angle Transformation

Simplifying with Euler's Identity

Squaring Both Sides

Connecting to the Product of Roots

Final Substitution and Proof

Summary of Key Concepts

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Roots

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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