Animated Solution for Mathematics - Complex Numbers: Let $S=\{a+b\sqrt{2}:a,b\in \mathbb{Z}\}$, $T_1=\{(-1+\sqrt{2}{)}^n:n\in \mathbb{Z}\}$, and $T_2=\{(1+\sqrt{2}{)}^n:n\in \mathbb{N}\}$. Then which of the following statements is(are) TRUE ?

Welcome, fellow traveler, to a beautiful exploration of number theory and complex analysis. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure known as $S=\{a+b\sqrt{2}:a,b\in \mathbb{Z}\}$.

Imagine this set as a playground where every number is built from the bedrock of integers and the mysterious, irrational $\sqrt{2}$.

We begin by defining our players. We have $S$, our primary domain. Then we have $T_1=\{(-1+\sqrt{2}{)}^n:n\in \mathbb{Z}\}$ and $T_2=\{(1+\sqrt{2}{)}^n:n\in \mathbb{N}\}$.

The question asks us to verify if these sets live within $S$. For any integer $k$, we can trivially write it as $k+0\sqrt{2}$, which fits the definition of $S$.

But what about the powers of $(1+\sqrt{2})$? When we expand $(1+\sqrt{2}{)}^n$ using the Binomial Theorem, we get:

Observe the magic: even powers of $\sqrt{2}$ become integers, and odd powers become an integer multiplied by $\sqrt{2}$. Thus, the entire sum collapses into the form $A+B\sqrt{2}$.

This confirms that $T_2$ is a subset of $S$. For $T_1$, the negative powers might look intimidating, but remember the conjugate property: $(\sqrt{2}-1)(\sqrt{2}+1)=1$.

This allows us to flip negative powers into positive ones, ensuring they too remain within the safe confines of $S$. Thus, the statement $\mathbb{Z}\cup T_1\cup T_2\subset S$ is undeniably TRUE.

Now, let us look at $T_1\cap (0,\frac{1}{2024})$. We are dealing with powers of $(\sqrt{2}-1)$.

Since $\sqrt{2}\approx 1.414$, our base is approximately $0.414$. When you repeatedly multiply a number between $0$ and $1$ by itself, it shrinks toward zero.

It doesn't just get small; it gets infinitely small. Therefore, for a sufficiently large $n$, the value of $(\sqrt{2}-1{)}^n$ will eventually fall into the tiny window $(0,\frac{1}{2024})$.

Because the intersection is not empty, the statement claiming it is empty is FALSE.

Conversely, consider $T_2=\{(1+\sqrt{2}{)}^n:n\in \mathbb{N}\}$. Here, our base is $1+\sqrt{2}\approx 2.414$.

This is greater than $1$. As we raise this to higher powers, the sequence grows without bound.

It will eventually soar past $2024$ and head toward infinity. Thus, $T_2\cap (2024,\infty )$ is definitely not empty. Statement C is TRUE.

Complex Numbers

Finally, we arrive at the most poetic part of our journey: the complex number $z=\cos (\pi (a+b\sqrt{2}))+i\sin (\pi (a+b\sqrt{2}))$. We are told this must be an integer.

Recall that any complex number $z$ on the unit circle has a magnitude of $1$. The only integers on the unit circle are $1$ and $-1$.

For $z$ to be $1$ or $-1$, the imaginary part must vanish:

This implies that the argument $\pi (a+b\sqrt{2})$ must be an integer multiple of $\pi$. Let this be $k\pi$.

Dividing by $\pi$, we get $a+b\sqrt{2}=k$. Rearranging gives $b\sqrt{2}=k-a$.

Since $k$ and $a$ are integers, their difference is an integer. If $b$ were anything other than zero, we would be claiming that $\sqrt{2}$ is rational, which we know is impossible.

Therefore, $b$ must be $0$. When $b=0$, $z=\cos (a\pi )=(-1{)}^a$, which is indeed an integer. Statement D is TRUE.