How does one transform the left side into the right side?
$$ (a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 $$
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Sign up to join this communityHow does one transform the left side into the right side?
$$ (a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 $$
There is a more conceptual explanation.
The complex conjugation is an automorphism of the field $\mathbb{C}$ of complex numbers (easy to see for any construction of $\mathbb{C}$). It follows that the norm function $N : \mathbb{C} \to \mathbb{R}, z \mapsto z \cdot \overline{z}$ is also multiplicative, i.e. satisfies $N(z z')=N(z) N(z')$. When $z=a+ib$ and $z'=c+id$, this means $(ad-bd)^2+(ad+bc)^2=(a^2+b^2)(c^2+d^2)$.
There is also such a formula for sums of four squares, using the quaternions $\mathbb{H}$, and the octonions $\mathbb{O}$ give a formula for sums of eight squares. But there is no such formula for sums of three squares, which corresponds to the fact there is no $3$-dimensional real normed algebra.
Use the method of foil on the right hand side and then foil the left hand side. It's easier doing right hand side to the left hand side.
$$(ac-bd)^2=(a^2c^2-2abcd+b^2d^2)$$ $$(ad+bc)^2=(a^2d^2+2abcd+b^2c^2)$$
Then simplify.
For the right hand side:
$$(ac-bd)^2+(ad+bc)^2= (a^2c^2-2abcd-b^2c^2)+(a^2d^2+2abcd+b^2c^2)$$ $$=a^2c^2+a^2d^2+b^2d^2+b^2c^2=RHS$$
Then take the left hand side and foil. $$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=LHS$$
Therefore
$RHS=LHS$ Does this make sense?
In the spirit of a more enlightening but less technical answer, we can think of the following.
Let $z=(a,b)$ and $w=(c,d)$ be two complex numbers. Then $|z|^2=a^2+b^2$ and $|w|^2=c^2+d^2$, while $|z\cdot w|^2=(ac-bd)^2+(ad+bc)^2$. So what we want to show is that for complex numbers $z,w$ $$|z|^2\cdot |w|^2=|z\cdot w|^2$$
But note that for any complex number, we have $|w|^2=w\cdot \bar w$. The above thus boils down to $$z\bar zw\bar w=zw\overline{zw}$$
Since $\Bbb C$ is commutative, all we need to show is that $\overline {zw}=\bar z\bar w$. Can you do that?
In fact, as Martin said, $z\mapsto \bar z$ is an field automorphism of $\Bbb C$, which means that $$\bar 1 =1\;\;, \bar 0=0$$ $$\overline{z+w}=\bar z+\bar w$$ and $$\overline{z\cdot w}=\bar z\cdot \bar w$$
Basically, it gives a bijection of $\Bbb C$ to itself which preserves the structure of $\Bbb C$: both identity elements are fixed, and the image of the sum or product of any numbers is the sum or product of the images.
$$(a^2+b^2)(c^2+d^2)$$ $$=a^2.c^2+a^2.d^2+b^2.c^2+b^2.d^2$$ $$=a^2.c^2+b^2.d^2+a^2.d^2+b^2.c^2$$ $$=a^2.c^2-2a.b.c.d+b^2.d^2+a^2.d^2+2a.b.c.d+b^2.c^2$$ $$=(ac)^2 - 2.(ac).(bd)+(bd)^2 + (ad)^2 + 2(ad)(bc) + (bc)^2$$ $$=(ac - bd)^2+(ad+bc)^2$$ $$=R.H.S$$
The most straighforward way is to transform the right side into the left side rather than the left side into the right side.
Expand the left hand side, you get $$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$ Add and substract $2abcd$ $$a^2c^2+a^2d^2+b^2c^2+b^2d^2=(a^2c^2-2abcd+b^2d^2)+(a^2d^2+2abcd+b^2c^2)$$ Complete the square, you can get $$(a^2c^2-2abcd+b^2d^2)+(a^2d^2+2abcd+b^2c^2)=(ac-bd)^2+(ad+bc)^2$$ Therefore, $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$
$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$
If $(a^2+b^2)\ne 0$ and $(c^2+d^2)\ne 0$ (If either of them is $0$ then the statement is vacuously true).
Let $\sin \alpha =\displaystyle \frac{a}{\sqrt{a^2+b^2}}\Rightarrow \cos \alpha\displaystyle \frac{b}{\sqrt{a^2+b^2}}$ and $\sin \beta =\displaystyle \frac{c}{\sqrt{c^2+d^2}}\Rightarrow \cos \beta\displaystyle \frac{d}{\sqrt{c^2+d^2}}$
So we have ,
$\displaystyle \frac{(ac-bd)^2 + (ad+bc)^2}{(a^2+b^2)(c^2+d^2)}=(-\cos (\alpha+\beta))^2+(\sin(\alpha+\beta))^2 =1$
We are done.