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

Question

How does one transform the left side into the right side?

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

Answer 1

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.

Answer 2

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?

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

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.