If $a + ib = \dfrac{(x + i)^{2}}{2x^{2} + 1}$, prove that $a^{2} + b^{2} = \dfrac{(x^{2} + 1)^{2}}{(2x^{2} + 1)^{2}}$.

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> [!TLDR]
> The modulus-squared of a complex number equals the product of the number and its conjugate; applying this to $a + ib = \frac{(x+i)^2}{2x^2+1}$ immediately gives $a^2 + b^2 = \left|\frac{(x+i)^2}{2x^2+1}\right|^2 = \frac{(x^2+1)^2}{(2x^2+1)^2}$.

The heart of this problem is recognizing what $a^2 + b^2$ represents. For any complex number $z = a + ib$, the quantity $a^2 + b^2$ is precisely $|z|^2$, the square of its modulus. Instead of expanding $(x+i)^2$ and separating real and imaginary parts—which works but is tedious—we can use the fundamental property that the modulus of a quotient is the quotient of the moduli.

The modulus of a complex number measures its distance from the origin, and crucially, $|z|^2 = z \cdot \overline{z}$. For a quotient, we have $\left|\frac{w}{v}\right| = \frac{|w|}{|v|}$, so squaring both sides gives us a clean path forward.

**Step-by-step proof:**

1. **Identify what we need to find.**  
   We're given $a + ib = \frac{(x+i)^2}{2x^2+1}$ and need to prove $a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2}$.  
   Since $a^2 + b^2 = |a+ib|^2$, we need to find the modulus-squared of the right-hand side.

2. **Apply the modulus to both sides.**  
   Taking the modulus of both sides:
   $$|a + ib| = \left|\frac{(x+i)^2}{2x^2+1}\right|$$

3. **Use the quotient property of modulus.**  
   The modulus of a quotient is the quotient of moduli:
   $$|a + ib| = \frac{|(x+i)^2|}{|2x^2+1|}$$

4. **Simplify the numerator using $|z^2| = |z|^2$.**  
   For any complex number $z$, we have $|z^n| = |z|^n$. Therefore:
   $$|(x+i)^2| = |x+i|^2$$ …

Q.If $a + ib = \dfrac{(x + i)^{2}}{2x^{2} + 1}$ , prove that $a^{2} + b^{2} = \dfrac{(x^{2} + 1)^{2}}{(2x^{2} + 1)^{2}}$ .

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Q.If $a + ib = \dfrac{(x + i)^{2}}{2x^{2} + 1}$ , prove that $a^{2} + b^{2} = \dfrac{(x^{2} + 1)^{2}}{(2x^{2} + 1)^{2}}$ .

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