Let S = {z ∈ C - {i, 2i} : (z² + 8iz - 15)/(z² - 3iz - 2) ∈ R}. If α - (13/11)i ∈ S, α ∈ R - {0}, th

Q: Wait, why do we use $BC - AD = 0$ instead of just setting the imaginary part of the fraction to zero directly?

Great question! While setting the imaginary part to zero is correct, doing it directly involves the denominator's magnitude squared, which is messy. Multiplying by the conjugate $\frac{C-iD}{C-iD}$ makes the denominator real, allowing us to focus solely on the numerator's imaginary part: $BC - AD = 0$. It is a much cleaner path to the solution.

Q: Why can we just divide by $x$ in the middle of the calculation?

We are given that $\alpha \in \mathbb{R} - \{0\}$, and since $z = \alpha - \frac{13}{11}i$, we know $x = \alpha$, so $x \neq 0$. In algebra, dividing by a variable is dangerous unless you are certain it is non-zero. Here, the problem statement explicitly protects us, so we can simplify without losing any solutions.

Q: Is there a geometric way to solve this without all this algebra?

Yes, you could interpret the expression as a ratio of two complex numbers and use the property that $\arg(z_1/z_2) = 0$ or $\pi$. This means the points $z_1, z_2, i, 2i$ are concyclic or collinear. However, the algebraic method is robust and less prone to geometric misinterpretation in the heat of an exam. Stick to the algebra for this one!

Question

Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}\}$. If $\alpha - \frac{13}{11}i \in S, \alpha \in \mathbb{R} - \{0\}$, then $242\alpha^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to peel back the layers of a problem that might look like a daunting algebraic mess, but is actually a beautiful geometric dance.

Imagine you are standing on the complex plane, looking at a set $S$ defined by a rational expression. We are told that for any $z \in S$, the expression

$$W = \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2}$$

must be purely real. This is our starting point.

### The Algebraic Grind: Deconstructing the Beast

To find the locus of $z$, the standard approach is to substitute $z = x + iy$, where $x$ and $y$ are real numbers.

For a complex number to be real, its imaginary part must be zero. If we write $W = \frac{A + iB}{C + iD}$, the condition for $W$ to be real is equivalent to $BC - AD = 0$. This is the **'Golden Key'** that simplifies our path.

### Expanding the Numerator and Denominator

Let's break down the numerator $z^2 + 8iz - 15$. Expanding this, we get:

$$(x^2 - y^2 - 8y - 15) + i(2xy + 8x)$$

We define the real part as $A = x^2 - y^2 - 8y - 15$ and the imaginary part as $B = 2xy + 8x$.

Now, for the denominator $z^2 - 3iz - 2$, we get:

$$(x^2 - y^2 + 3y - 2) + i(2xy - 3x)$$

We define the real part as $C = x^2 - y^2 + 3y - 2$ and the imaginary part as $D = 2xy - 3x$.

### The Golden Condition

Now, we apply our condition $BC - AD = 0$. Substituting our expressions for $A, B, C,$ and $D$, we get:

$$(2xy + 8x)(x^2 - y^2 + 3y - 2) - (x^2 - y^2 - 8y - 15)(2xy - 3x) = 0$$

Notice that both $B$ and $D$ contain a factor of $x$. Since $\alpha 
eq 0$, we know $x 
eq 0$, so we can safely divide the entire equation by $x$. This leaves us with:

$$(2y + 8)(x^2 - y^2 + 3y - 2) - (2y - 3)(x^2 - y^2 - 8y - 15) = 0$$

### The Geometric Revelation

Now, we expand carefully. After distributing the terms and combining like terms, the $x^2$ and $y^2$ terms simplify beautifully.

We are left with the equation:

$$11x^2 + 11y^2 + 26y - 61 = 0$$

Look at that! It is the equation of a circle. We have successfully translated an abstract complex condition into a concrete geometric shape.

### The Final Victory

The problem tells us that $z = \alpha - \frac{13}{11}i$ lies on this circle. This means its coordinates $(\alpha, -\frac{13}{11})$ must satisfy our circle equation.

Substituting $x = \alpha$ and $y = -\frac{13}{11}$ into the equation, we get:

$$11\alpha^2 + 11\left(-\frac{13}{11}ight)^2 + 26\left(-\frac{13}{11}ight) - 61 = 0$$

Simplifying this, we find:

$$11\alpha^2 + \frac{169}{11} - \frac{338}{11} - 61 = 0$$

$$11\alpha^2 - \frac{169}{11} - 61 = 0 \Rightarrow 11\alpha^2 = \frac{169 + 671}{11} = \frac{840}{11}$$

Multiplying by $11$, we get $121\alpha^2 = 840$. The question asks for $242\alpha^2$, which is exactly $2 	imes 121\alpha^2$.

Thus, $2 	imes 840 = 1680$. We have arrived at the finish line. The final answer is **1680**.

Animated Solution for Mathematics - Complex Numbers: Let $S = {z \in C - {i, 2 i} : \frac{z ^{2} + 8 i z - 15}{z ^{2} - 3 i z - 2} \in R}$ . If $α - \frac{13}{11} i \in S, α \in R - {0}$ , then $242 α^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Defining the Set $S$

Substituting $z = x + i y$

Expanding the Numerator

Expanding the Denominator

The Condition for Real Value

Setting up the Locus Equation

Expanding the First Term

Expanding the Second Term

Combining and Simplifying

Substituting the Point $(α, - \frac{13}{11})$

Solving for $11 α^{2}$

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Algebraic Grind

Expanding the Numerator and Denominator

The Golden Condition

The Geometric Revelation

The Final Victory

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z∈C−{i,2i}:z2−3iz−2z2+8iz−15​∈R}. If α−1113​i∈S,α∈R−{0}, then 242α2 is equal to