Let z1, z2 and z3 be three complex numbers on the circle |z| = 1 with arg(z1) = -π/4, arg(z2) = 0 an

Q: Wait, why can't we just use the triangle inequality to find the modulus? It seems faster.

The triangle inequality gives you a bound, not an exact value! It tells you the maximum or minimum possible, but here we need the exact value of the squared modulus. We must calculate the complex number $S$ explicitly to find the exact coefficients $\alpha$ and $\beta$.

Q: Why is the conjugate of $z_2$ just 1? I thought conjugates always flip the sign of the imaginary part.

You are absolutely right! But remember, $z_2$ has an argument of $0$, meaning it lies on the positive real axis. Its imaginary part is $0$. So, flipping the sign of $0$ still gives you $0$. It's a real number, and the conjugate of a real number is itself.

Q: How do we know we can compare $\alpha$ and $\beta$ directly? What if there are other terms?

Great question! Since $\alpha$ and $\beta$ are integers, and $\sqrt{2}$ is irrational, the representation $\alpha + \beta\sqrt{2}$ is unique. This is a fundamental property of the field $\mathbb{Q}(\sqrt{2})$. You can safely equate the rational and irrational parts.

Question

Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $arg(z_{1})=\frac{-\pi}{4}, arg(z_{2})=0$ and $arg(z_{3})=\frac{\pi}{4}$. If $|z_{1}\overline{z}_{2}+z_{2}\overline{z}_{3}+z_{3}\overline{z}_{1}|^{2}=\alpha+\beta\sqrt{2}, \alpha, \beta\in Z,$ then the value of $\alpha^{2}+\beta^{2}$ is:

Options:

(A) 24

(B) 29

(C) 41

(D) 31

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Unit Circle

Welcome, fellow traveler, to the beautiful world of complex numbers. Today, we are not just solving an equation; we are exploring the geometry of the unit circle.

Imagine a circle centered at the origin $(0,0)$ with a radius of exactly $1$. This is our playground. Every complex number $z$ on this circle satisfies $|z|=1$, which means we can represent any point on this boundary using Euler's formula: $z = e^{i	heta}$.

By converting the geometric constraints into algebraic forms, we turn a daunting problem into a simple dance of exponents.

### Locating the Points

We are given three points: $z_1, z_2,$ and $z_3$. Their arguments are given as $-\frac{\pi}{4}$, $0$, and $\frac{\pi}{4}$ respectively.

$z_2$ sits proudly on the positive real axis because its argument is $0$. $z_3$ is at $\frac{\pi}{4}$, pointing into the first quadrant at a $45^{\circ}$ angle. $z_1$ is at $-\frac{\pi}{4}$, mirroring $z_3$ in the fourth quadrant.

Using Euler's form, we write:
$$z_2 = e^{i(0)} = 1$$
$$z_3 = e^{i\pi/4} = \cos\left(\frac{\pi}{4}ight) + i\sin\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$$
$$z_1 = e^{-i\pi/4} = \cos\left(\frac{\pi}{4}ight) - i\sin\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$$

See how symmetrical they are? This symmetry is a hint that things will simplify beautifully.

### The Algebraic Grind

Now, let's tackle the expression $S = z_1\overline{z}_2 + z_2\overline{z}_3 + z_3\overline{z}_1$. We take it term by term.

First, $z_1\overline{z}_2$. Since $z_2 = 1$, its conjugate is also $1$. Thus, $z_1\overline{z}_2 = z_1 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$.

Second, $z_2\overline{z}_3$. Since $z_2 = 1$, this is just the conjugate of $z_3$. Flipping the sign of the imaginary part of $z_3$, we get $\overline{z}_3 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$. Notice that $z_1\overline{z}_2$ and $z_2\overline{z}_3$ are identical.

Finally, the third term: $z_3\overline{z}_1$. This is:
$$\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}ight)\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}ight) = \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}ight)^2$$

Expanding this using $(a+b)^2 = a^2 + b^2 + 2ab$, we get:
$$\frac{1}{2} - \frac{1}{2} + i = i$$
The real parts cancel out, leaving us with just $i$.

### The Grand Finale

We sum these up:
$$S = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}ight) + \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}ight) + i$$

Combining the real parts, we get $\frac{2}{\sqrt{2}} = \sqrt{2}$. Combining the imaginary parts, we get:
$$-\frac{i}{\sqrt{2}} - \frac{i}{\sqrt{2}} + i = -i\sqrt{2} + i = i(1 - \sqrt{2})$$

So, $S = \sqrt{2} + i(1 - \sqrt{2})$. The problem asks for $|S|^2$, which is the real part squared plus the imaginary part squared:
$$|S|^2 = (\sqrt{2})^2 + (1 - \sqrt{2})^2 = 2 + (1 + 2 - 2\sqrt{2}) = 5 - 2\sqrt{2}$$

Comparing this to $\alpha + \beta\sqrt{2}$, we find $\alpha = 5$ and $\beta = -2$. Finally, **$\alpha^2 + \beta^2 = 25 + 4 = 29$**.

We have arrived at the destination. The beauty of this problem lies not just in the answer, but in how the complex numbers, initially scattered on the circle, collapsed into such a clean, elegant result.

Select Answer:

Visualized Solution

Visualizing the Unit Circle

Locating $z_{1}, z_{2}, z_{3}$

Euler's Form Representation

Calculating $z_{1} \overline{z}_{2}$

Calculating $z_{2} \overline{z}_{3}$

Calculating $z_{3} \overline{z}_{1}$ (Setup)

Calculating $z_{3} \overline{z}_{1}$ (Execution)

Summing the Terms

Calculating Squared Modulus $∣ S ∣^{2}$

Finding $α$ and $β$

Final Calculation: $α^{2} + β^{2}$

The Sigma Insight: Argand Plane and Polar Representation

Analyzing the Geometry of the Unit Circle

Locating the Points

The Algebraic Grind

The Grand Finale

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Select Answer:

Visualized Solution

Visualizing the Unit Circle

Locating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​,z2​,z3​

Euler's Form Representation

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​z2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​z3​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z3​z1​ (Setup)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z3​z1​ (Execution)

Summing the Terms

Calculating Squared Modulus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣S∣2

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2