# How do I transform the left side into the right side of this equation?

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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Where is the equation? –  Patrick Li Mar 15 '13 at 2:31
Image was broken, it's fixed now –  Dan Webster Mar 15 '13 at 2:32
I got rid of the image completely. If anyone wants to see it, look at the edit history. –  Michael Hardy Mar 15 '13 at 2:53

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.

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Thank you for this more gentle version of my answer. –  Martin Brandenburg Mar 15 '13 at 12:16

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.

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How (un)Helpful for the OP! –  amWhy Mar 15 '13 at 2:56
I promote this point of view here because it might be interesting and perhaps inspiring for some readers (which of course are more than just the OP). If you don't like my answer, feel free to downvote it. But please don't judge what is helpful for others. –  Martin Brandenburg Mar 15 '13 at 2:58
But more importantly I would like to point out that mathematics, even this simple equation, is much more than just doing stupid computations, as in the other answers. They also don't offer any explanation why we can write the product of two sums of two squares as a sum of two squares and how it looks like explicitly. This is best done with the help of complex numbers (or by the genius of Fibonacci?). –  Martin Brandenburg Mar 15 '13 at 3:01
Fair enough. But in so doing, please clarify that "mathematics, even this simple equation, is much more than...". There's a time and place for "pontificating." And No, I don't have any intention of downvoting. It is not a downvote candidate. –  amWhy Mar 15 '13 at 3:10

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?

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$$(a^2+b^2)(c^2+d^2)$$ $$=a^2.c^2+a^2.d^2+b^2.c^2+b^2.d^2$$ $$=a^2.c^2+b^2.d^2+a^2.d^2+b^2.c^2$$ $$=a^2.c^2-2a.b.c.d+b^2.d^2+a^2.d^2+2a.b.c.d+b^2.c^2$$ $$=(ac)^2 - 2.(ac).(bd)+(bd)^2 + (ad)^2 + 2(ad)(bc) + (bc)^2$$ $$=(ac - bd)^2+(ad+bc)^2$$ $$=R.H.S$$

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The most straighforward way is to transform the right side into the left side rather than the left side into the right side.

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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, $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$

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$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$

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.

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