If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$, then show that $(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2}) = A^{2} + B^{2}$.

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> [!TLDR]
> The product of complex numbers multiplies their moduli, and the modulus squared of a complex number equals the sum of squares of its real and imaginary parts. So the given identity follows directly from $|z_1 z_2 z_3 z_4|^2 = |z_1|^2 |z_2|^2 |z_3|^2 |z_4|^2$.

The problem asks you to connect a product of four complex numbers with the product of their individual "lengths." This is a classic application of the modulus (or absolute value) of a complex number.

For any complex number $z = x + iy$, its modulus is defined as $|z| = \sqrt{x^2 + y^2}$. Squaring that gives $|z|^2 = x^2 + y^2$. That's exactly the form $(a^2 + b^2)$ you see on the left side of the target equation — it's $|a+ib|^2$.

Now, the key property: when you multiply complex numbers, their moduli multiply. That is, $|z_1 z_2| = |z_1| |z_2|$. This holds for any number of factors: $|z_1 z_2 \cdots z_n| = |z_1| |z_2| \cdots |z_n|$.

So here, let:
- $z_1 = a + ib$
- $z_2 = c + id$
- $z_3 = e + if$
- $z_4 = g + ih$

Their product is given as $A + iB$. Therefore:

$$|z_1 z_2 z_3 z_4| = |A + iB|$$

But by the product property:

$$|z_1 z_2 z_3 z_4| = |z_1| |z_2| |z_3| |z_4|$$

Now square both sides:

$$|z_1|^2 |z_2|^2 |z_3|^2 |z_4|^2 = |A + iB|^2$$

And since $|x+iy|^2 = x^2 + y^2$ for any complex number, we get:

$$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$$

That's the entire proof — it's really just one idea applied cleanly. …

Q.If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ , then show that $(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2}) = A^{2} + B^{2}$ .

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Q.If (a+ib)(c+id)(e+if)(g+ih)=A+iB(a + ib)(c + id)(e + if)(g + ih) = A + iB(a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2}) = A^{2} + B^{2}(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2.

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Q.If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ , then show that $(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2}) = A^{2} + B^{2}$ .

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