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The Pythagorean theorem written as $$ a^2 + b^2 = c^2 $$ has the simply geometric meaning that the sum of the areas of the two squares on the legs ($a$ and $b$) equals the area of the square on the hypotenuse $c$

enter image description here

But algebraically the Pythagorean theorem can also be written using complex numbers like $$ (a + ib)(a - ib) = c^2 $$ Here instead of a sum, we have a product of two complex quantities. Can the latter equation also be interpreted geometrically somehow?

2 Answers 2

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Amazingly, you can interpret this. Notice that we have

$$a+ib=\overline{a-ib}$$

That is, they are conjugates. Thus,

$$z\overline z=|z|^2$$

Or,

$$(a+ib)(a-ib)=|a+ib|^2$$

The absolute value of a complex number is it's distance from $0$, or, graphically,

enter image description here

That is,

$$c=|a+bi|$$

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    Since $c$ is a real number shouldn't the point z be somehow on the Re axis, too? – 2017-02-04
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    @asmaier No. $|\cdot|$ always makes things a positive real number, and I assumed $c>0$. – 2017-02-04
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    Would $c=|a-bi|$ be correct, too? – 2017-02-04
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    @asmaier Yes. If you think about that, that would just be the graph flipped upside down (referring to image in my answer) – 2017-02-04
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maybe worth noting that if $[a,b,r]$ is a primitive Pythagorean triple, with hypotenuse the odd positive integer $r$, and $a$ the other side of odd length, then the Gaussian integer $a \pm ib$ has "exact" square roots: $$ \sqrt{a\pm ib} = \pm \bigg(\sqrt{\frac{r+a}2} \pm i\sqrt{\frac{r-a}2} \bigg) $$ since the quantities in the surds are perfect squares.

conversely, by squaring the Gaussian integer $(m+in)$, with $m \gt n \gt 0$ we obtain $(m^2-n^2)+2imn$, yielding the Pythagorean triad $(m^2-n^2,2mn,m^2+n^2)$