Let w = zz̄ + k1z + k2iz + λ(1 + i), k1, k2 ∈ R. Let Re(w) = 0 be the circle C of radius 1 in the fi

Q: Wait, why is the real part of $w$ a circle? I thought complex numbers were just points.

That is a fantastic question! Think of the complex plane as a coordinate plane where $z = x + iy$. When we set the real part of an expression involving $z$ to zero, we are essentially imposing a constraint on the coordinates $(x, y)$. In this case, the expression $x^2 + y^2 + k_1x - k_2y + \lambda = 0$ is the classic algebraic form of a circle. It is not just a point; it is the set of all points $(x, y)$ that satisfy this specific geometric condition.

Q: How do I know the center of the circle is $(1, 2)$?

Great observation. The problem tells us two vital clues: the circle has a radius of 1, it touches the $y$-axis, and it touches the line $y = 1$ in the first quadrant. If a circle of radius 1 touches the $y$-axis ($x=0$), its center must be 1 unit away from the $y$-axis, so the $x$-coordinate is 1. If it touches the line $y=1$ from above, its center must be 1 unit above that line, so the $y$-coordinate is $1 + 1 = 2$. Thus, the center is $(1, 2)$.

Q: Can I use the distance formula directly without finding the coordinates of $A$ and $B$?

You could, but it is much more elegant to find the coordinates. Since the line $y = 2x + 2$ intersects the circle, substituting $y$ into the circle equation gives you a quadratic in $x$. Solving this quadratic gives you the $x$-coordinates of the intersection points directly. Once you have $x$, finding $y$ is trivial. It is a very robust and reliable method!

Question

Let $w = z\bar{z} + k_1z + k_2iz + \lambda(1 + i), k_1, k_2 \in \mathbb{R}$. Let $	ext{Re}(w) = 0$ be the circle $C$ of radius 1 in the first quadrant touching the line $y = 1$ and the $y$-axis. If the curve $	ext{Im}(w) = 0$ intersects $C$ at $A$ and $B$, then $30(AB)^2$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers

Welcome, fellow traveler of the mathematical landscape. Today, we are going to peel back the layers of a problem that might look like a daunting algebraic mess, but is, in reality, a beautiful dance between complex numbers and coordinate geometry.

When you see an expression like $w = z\bar{z} + k_1z + k_2iz + \lambda(1 + i)$, do not panic. Instead, see it as a hidden map waiting to be decoded.

### Phase 1: The Expansion

We begin by grounding our complex variable $z$ in the familiar territory of the Cartesian plane. Let $z = x + iy$. This implies that its conjugate, $\bar{z}$, is $x - iy$.

Now, we substitute these into our expression for $w$. As we expand $w = (x + iy)(x - iy) + k_1(x + iy) + k_2i(x + iy) + \lambda(1 + i)$, we are essentially separating the world into two realms: the real and the imaginary.

The term $z\bar{z}$ becomes $x^2 + y^2$, a classic component of circular geometry. By grouping the terms without $i$ and those with $i$, we reveal the structure of the problem:

$$	ext{Re}(w) = x^2 + y^2 + k_1x - k_2y + \lambda = 0$$

$$	ext{Im}(w) = k_1y + k_2x + \lambda = 0$$

### Phase 2: The Geometric Revelation

Now, look at the real part. It is the equation of a circle! The problem gives us the radius $r = 1$ and two crucial tangency conditions.

It touches the $y$-axis, meaning the distance from the center to the $y$-axis is the radius, so the $x$-coordinate of the center is $1$. It also touches the line $y = 1$ in the first quadrant, which means the center must be $1$ unit above that line, placing the $y$-coordinate at $1 + 1 = 2$.

Our circle is centered at $(1, 2)$ with radius $1$. The standard equation is:

$$(x - 1)^2 + (y - 2)^2 = 1^2$$

This expands to:

$$x^2 + y^2 - 2x - 4y + 4 = 0$$

### Phase 3: The Algebraic Bridge

This is the moment of truth. We compare our expanded circle equation $x^2 + y^2 - 2x - 4y + 4 = 0$ with our derived expression $	ext{Re}(w) = x^2 + y^2 + k_1x - k_2y + \lambda = 0$.

By matching the coefficients, the path forward is illuminated: $k_1 = -2$, $-k_2 = -4$ (so $k_2 = 4$), and $\lambda = 4$. We have successfully bridged the gap between the complex expression and the geometric reality.

### Phase 4: The Intersection

With our constants in hand, the imaginary part $	ext{Im}(w) = 0$ becomes a simple linear equation: $-2y + 4x + 4 = 0$, which simplifies beautifully to $y = 2x + 2$.

This line cuts through our circle. To find the points of intersection $A$ and $B$, we substitute $y = 2x + 2$ into the circle equation $(x - 1)^2 + (y - 2)^2 = 1$.

The math simplifies with satisfying elegance:

$$(x - 1)^2 + (2x + 2 - 2)^2 = 1$$

$$(x - 1)^2 + 4x^2 = 1$$

Expanding this gives $5x^2 - 2x = 0$, yielding $x = 0$ and $x = \frac{2}{5}$.

### Phase 5: The Final Calculation

We are almost there. For $x = 0$, $y = 2$, so $A = (0, 2)$. For $x = \frac{2}{5}$, $y = 2(\frac{2}{5}) + 2 = \frac{14}{5}$, so $B = (\frac{2}{5}, \frac{14}{5})$.

The square of the distance $(AB)^2$ is:

$$(AB)^2 = \left(\frac{2}{5} - 0ight)^2 + \left(\frac{14}{5} - 2ight)^2 = \frac{4}{25} + \frac{16}{25} = \frac{20}{25} = \frac{4}{5}$$

Finally, the problem asks for $30(AB)^2$, which is:

$$30 	imes \frac{4}{5} = \mathbf{24}$$

What a journey! We started with a complex expression and ended with a precise numerical value, having navigated through geometry and algebra. Keep this spirit of inquiry alive; every problem is just a story waiting for you to tell it.

Enter Numerical Value:

Visualized Solution

Expansion of $w$

Separating Real and Imaginary Parts

Analyzing the Circle $C$

Equation of Circle $C$

Finding $k_{1}, k_{2},$ and $λ$

The Curve $Im (w) = 0$

Finding Intersection Points $A$ and $B$

Coordinates of $A$ and $B$

Calculating Distance Squared $(A B)^{2}$

Final Answer Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Enter Numerical Value:

Visualized Solution

Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

Separating Real and Imaginary Parts

Analyzing the Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Equation of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k1​,k2​, and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

The Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Im(w)=0

Finding Intersection Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating Distance Squared .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (AB)2