Question

Consider the curves A, B, C, D defined as

$A = \left\{ (x,y) : x^2 - y^2 = \frac{x}{x^2 + y^2} \right\}$

$B = \left\{ (x,y) : 2xy + \frac{y}{x^2 + y^2} = 3 \right\}$

$C = \left\{ (x,y) : x^3 - 3xy^2 + 3y = 1 \right\}$

$D = \left\{ (x,y) : 3x^2y - 3x - y^3 = 0 \right\}$

Then $\frac{n(A \cap B)}{n(C \cap D)} =$ _____. $(x, y \in R$ and n(S) represents number of elements in S)

Answer 1

1

Answer 2

To solve this problem, we will use the properties of complex numbers. Let $z = x + iy$, where $x, y \in \mathbb{R}$.

Then, we have the following relations:

1. $z^2 = (x+iy)^2 = x^2 - y^2 + 2xyi$
2. $z^3 = (x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = x^3 - 3xy^2 + i(3x^2y - y^3)$
3. $|z|^2 = x^2 + y^2$
4. $\frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$

Analyzing Curve A and B:

Curve A: $x^2 - y^2 = \frac{x}{x^2 + y^2}$

This equation can be written in terms of complex numbers as:

$\text{Re}(z^2) = \text{Re}\left(\frac{1}{z}\right)$

Curve B: $2xy + \frac{y}{x^2 + y^2} = 3$

This equation can be written in terms of complex numbers as:

$\text{Im}(z^2) - \text{Im}\left(\frac{1}{z}\right) = 3$ (Note: $\text{Im}(1/z) = -y/(x^2+y^2)$)

For a point $(x,y)$ to be in $A \cap B$, both conditions must be satisfied.

This implies:

$\text{Re}(z^2) - \text{Re}\left(\frac{1}{z}\right) = 0$

$\text{Im}(z^2) - \text{Im}\left(\frac{1}{z}\right) = 3$

Combining these two equations, we get:

$z^2 - \frac{1}{z} = (\text{Re}(z^2) - \text{Re}(1/z)) + i(\text{Im}(z^2) - \text{Im}(1/z))$

$z^2 - \frac{1}{z} = 0 + 3i$

$z^2 - \frac{1}{z} = 3i$

To eliminate the denominator, multiply by $z$ (note that $z=0$ would make $1/z$ undefined, so $z \ne 0$. Also, $z=0$ is not a solution to $z^2 - 1/z = 3i$ as $0 - \text{undefined} \ne 3i$):

$z^3 - 1 = 3iz$

Rearranging the terms, we get a cubic equation in $z$:

$z^3 - 3iz - 1 = 0$

Let $f(z) = z^3 - 3iz - 1$. To find the number of distinct roots, we check for repeated roots by finding the common roots of $f(z)=0$ and $f'(z)=0$.

$f'(z) = 3z^2 - 3i$

Set $f'(z) = 0$:

$3z^2 - 3i = 0 \implies z^2 = i$

The roots for $z^2 = i$ are $z = \pm \sqrt{i}$.

We know $\sqrt{i} = \sqrt{e^{i\pi/2}} = e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$.

Let $z_0 = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$.

Now, substitute $z_0$ into $f(z)$:

$f(z_0) = z_0^3 - 3iz_0 - 1$. Since $z_0^2 = i$, we have $z_0^3 = iz_0$.

$f(z_0) = iz_0 - 3iz_0 - 1 = -2iz_0 - 1$

$f(z_0) = -2i\left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) - 1 = -\sqrt{2}i - \sqrt{2}i^2 - 1 = -\sqrt{2}i + \sqrt{2} - 1$.

Since $f(z_0) \ne 0$, $z_0$ is not a root of $f(z)=0$.

Similarly, for $z_1 = -z_0$:

$f(z_1) = (-z_0)^3 - 3i(-z_0) - 1 = -z_0^3 + 3iz_0 - 1 = -iz_0 + 3iz_0 - 1 = 2iz_0 - 1$.

$f(z_1) = 2i\left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) - 1 = \sqrt{2}i + \sqrt{2}i^2 - 1 = \sqrt{2}i - \sqrt{2} - 1$.

Since $f(z_1) \ne 0$, $z_1$ is not a root of $f(z)=0$.

Since none of the roots of $f'(z)=0$ are roots of $f(z)=0$, the cubic equation $z^3 - 3iz - 1 = 0$ has 3 distinct roots. Each distinct complex root $z=x+iy$ corresponds to a unique point $(x,y)$ in the real plane.

Thus, $n(A \cap B) = 3$.

Analyzing Curve C and D:

Curve C: $x^3 - 3xy^2 + 3y = 1$

This equation can be written in terms of complex numbers as:

$\text{Re}(z^3) + 3y = 1$

Curve D: $3x^2y - 3x - y^3 = 0$

This equation can be written in terms of complex numbers as:

$\text{Im}(z^3) - 3x = 0$

For a point $(x,y)$ to be in $C \cap D$, both conditions must be satisfied.

This implies:

$\text{Re}(z^3) = 1 - 3y$

$\text{Im}(z^3) = 3x$

So, $z^3 = \text{Re}(z^3) + i\text{Im}(z^3) = (1 - 3y) + i(3x)$.

We want to express the right side in terms of $z = x+iy$.

Notice that $3ix - 3y = 3i(x+iy) = 3iz$.

So, $z^3 = 1 + (3ix - 3y) = 1 + 3i(x+iy) = 1 + 3iz$.

Rearranging the terms, we get:

$z^3 - 3iz - 1 = 0$

This is the exact same cubic equation as derived for $A \cap B$.

Since the equation is the same, and we have already established that it has 3 distinct roots, $n(C \cap D) = 3$.

Also, $(0,0)$ is not a solution for C, as $0-0+0 = 1$ is false. So $z=0$ is not a solution for $C \cap D$.

Calculating the Ratio:

We found $n(A \cap B) = 3$ and $n(C \cap D) = 3$.

Therefore, $\frac{n(A \cap B)}{n(C \cap D)} = \frac{3}{3} = 1$.

The final answer is $\boxed{1}$.

Explanation of the solution:

The problem involves finding the number of intersection points for two pairs of curves. By expressing the given Cartesian equations in terms of complex numbers $z = x+iy$, it was found that both $A \cap B$ and $C \cap D$ lead to the same cubic equation: $z^3 - 3iz - 1 = 0$.

Analysis of this cubic equation and its derivative revealed that it has no repeated roots, meaning it has 3 distinct complex roots. Each distinct complex root corresponds to a unique point $(x,y)$ in the real plane.

Therefore, $n(A \cap B) = 3$ and $n(C \cap D) = 3$. The required ratio is $3/3 = 1.