If α denotes the number of solutions of |1 - i|ˣ = 2ˣ and β = |z|/arg(z), where z = π/4(1 + i)⁴ ((1

Q: Wait, why is $\alpha = 1$? I thought exponential equations could have multiple solutions.

Great question! When we simplify $(\sqrt{2})^x = 2^x$ to $2^{x/2} = 2^x$, we are equating the exponents. The only value for which $x/2 = x$ is $x = 0$. Since there is only one value of $x$ that satisfies the equation, the number of solutions, $\alpha$, is indeed 1.

Q: Do I really need to rationalize those complex fractions? It looks like a nightmare.

Absolutely not! In JEE Advanced, if a calculation looks like a nightmare, there is almost always a hidden symmetry. Notice that $-i(\sqrt{\pi} + i) = -i\sqrt{\pi} - i^2 = 1 - i\sqrt{\pi}$. The numerator is just the denominator multiplied by $-i$. Spotting these patterns is the hallmark of a top-tier student.

Q: Why is the argument of $z = 2\pi i$ equal to $\pi/2$?

Think geometrically. A complex number $z = x + iy$ is a point $(x, y)$ in the Argand plane. Here, $z = 0 + 2\pi i$, which corresponds to the point $(0, 2\pi)$. This point lies on the positive imaginary axis. The angle it makes with the positive real axis is exactly $90^\circ$ or $\pi/2$ radians.

Question

If $\alpha$ denotes the number of solutions of $|1 - i|^x = 2^x$ and $\beta = \frac{|z|}{	ext{arg}(z)}$, where $z = \frac{\pi}{4}(1 + i)^4 (\frac{1 - \sqrt{\pi}i}{\sqrt{\pi} + i} + \frac{\sqrt{\pi} - i}{1 + \sqrt{\pi}i})$, $i = \sqrt{-1}$, then the distance of the point $(\alpha, \beta)$ from the line $4x - 3y = 7$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The detective's approach to this problem requires us to dismantle the complexity layer by layer. We are tasked with finding the distance of a point $(\alpha, \beta)$ from a specific line, but first, we must determine the values of $\alpha$ and $\beta$ through two distinct mathematical phases.

### Phase 1: The Exponential Mystery

We begin with the equation $|1 - i|^x = 2^x$. The modulus of a complex number $z = a + bi$ is defined as $|z| = \sqrt{a^2 + b^2}$.

For the term $|1 - i|$, we calculate:
$$|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$

Substituting this back into the original equation, we obtain:
$$(\sqrt{2})^x = 2^x$$

Since $\sqrt{2} = 2^{1/2}$, the equation becomes $2^{x/2} = 2^x$. Equating the exponents, we have $\frac{x}{2} = x$, which yields $x = 0$. Thus, the number of solutions is **$\alpha = 1$**.

### Phase 2: The Complex Labyrinth

Next, we simplify the expression for $z$. We first evaluate $(1 + i)^4$:
$$(1 + i)^2 = 1 + 2i + i^2 = 2i$$
$$(2i)^2 = 4i^2 = -4$$

Now, consider the fractions $\frac{1 - \sqrt{\pi}i}{\sqrt{\pi} + i}$ and $\frac{\sqrt{\pi} - i}{1 + \sqrt{\pi}i}$. By multiplying the denominator of the first fraction by $-i$, we get $-i\sqrt{\pi} - i^2 = 1 - i\sqrt{\pi}$, which matches the numerator. Thus, the first fraction simplifies to $-i$.

Applying the same logic to the second fraction, multiplying the denominator $(1 + \sqrt{\pi}i)$ by $-i$ yields $-i - \sqrt{\pi}i^2 = \sqrt{\pi} - i$. This also simplifies to $-i$. Substituting these into the expression for $z$:
$$z = \frac{\pi}{4} (-4) (-i - i)$$
$$z = -\pi (-2i) = 2\pi i$$

### Phase 3: The Geometric Finale

We have $z = 2\pi i$. The modulus is $|z| = 2\pi$, and the argument of a purely imaginary number on the positive imaginary axis is $	ext{arg}(z) = \frac{\pi}{2}$.

Calculating $\beta$:
$$\beta = \frac{|z|}{	ext{arg}(z)} = \frac{2\pi}{\pi/2} = 4$$

Our point is $(\alpha, \beta) = (1, 4)$. We now find the perpendicular distance from $(1, 4)$ to the line $4x - 3y - 7 = 0$ using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$:
$$d = \frac{|4(1) - 3(4) - 7|}{\sqrt{4^2 + (-3)^2}}$$
$$d = \frac{|4 - 12 - 7|}{\sqrt{16 + 9}} = \frac{|-15|}{5} = 3$$

The final distance is **$3$**.

Enter Numerical Value:

Visualized Solution

Problem Overview

Solving for $α$

Finding the Number of Solutions

Simplifying $(1 + i)^{4}$

Simplifying the Fractions

Simplifying the Second Fraction

Calculating the Final Value of $z$

Finding Modulus and Argument of $z$

Visualizing the Point and Line

Applying the Distance Formula

Final Calculation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ where $i = - 1$ , then $16 \cdot R e (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}}) \cdot I m (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}})$ is equal to

If for $z = α + i β$ , $∣ z + 2∣ = z + 4 (1 + i)$ , then $α + β$ and $α β$ are the roots of the equation

Find the real values of $x$ and $y$ for which the following equation is satisfied $\frac{( 1 + i ) x - 2 i}{3 + i} + \frac{( 2 - 3 i ) y + i}{3 - i} = i$ .

If $α, β \in R$ are such that $1 - 2 i$ (here $i^{2} = - 1$ ) is a root of $z^{2} + α z + β = 0$ , then $(α - β)$ is equal to:

Let $a, b, x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq = 0$ . If the complex number $z = x + i y$ satisfies $Im (\frac{a z + b}{z + 1}) = y$ , then which of the following is(are) possible value(s) of $x$ ?

If the least and the largest real values of $α$ , for which the equation $z + α ∣ z -1∣ + 2 i = 0$ ( $z \in C$ and $i = - 1$ ) has a solution, are p and q respectively; then $4 (p^{2} + q^{2})$ is equal to

Let $n$ denote the number of solutions of the equation $z^{2} + 3 \overset{z}{ˉ} = 0$ , where $z$ is a complex number. Then the value of $\sum_{k = 0}^{\infty} \frac{1}{n ^{k}}$ is equal to

Let $\overset{z}{ˉ}$ denote the complex conjugate of a complex number $z$ and let $i = - 1$ . In the set of complex numbers, the number of distinct roots of the equation $\overset{z}{ˉ} - z^{2} = i (\overset{z}{ˉ} + z^{2})$ is _____________.

If $(\frac{1 + i}{1 - i})^{x} = 1$ then