# Baby Rudin 1.14 is essentially Thales's theorem?

Exercise 1.14 in Baby Rudin asks to compute $$\mid 1+z\mid^2+\mid 1-z\mid^2$$ for (arbitrary) complex $z$ lying on the unit circle. This evaluates to $4$.

Consider the triangle $0,1+z,1-z$ and shift it to the left by $1$. We get a triangle inscribed in a circle with one of its sides lying on a diagonal. Conversely, if we start with any triangle inscribed in a circle lying on a diagonal we can embed the figure in $\mathbb{C}$, so that the image of the circle is the unit circle, and the vertices of the triangle are $-1,z,-z$ for some $z\in\mathbb{C}$. Now, shift the triangle to the right by 1. Two of its sides become $1+z$ and $1-z$. Our computation shows that this triangle is a right triangle. This result is known as the Thales's theorem.

Question 1. Is this a well-known proof of the Thales's theorem?

Question 2. Did the author actually mean this to be "discovered"? If yes, then should one expect to find a lot of such "purposefully hidden" stuff in Rudin's books?

• I conjecture that the intended proof was to write $|1+z|^2$ as $(1+z)(1+\bar z)$ and similarly for the other term, multiply everything out, and cancel some terms. Less elegant but correct would be to plug in $z=\cos\theta+i\sin\theta$ and calculate from there. I like your proof better, and it might be how Rudin came up with the exercise in the first place, but I'd be surprised if he intended typical readers to discover it. Jul 27, 2018 at 0:11
• I meant the proof of Thales's theorem using this identity with complex numbers, not the other way around. Though both statements seem to be equivalent. I'll edit the question to make it clearer. Jul 27, 2018 at 0:17
• @AndreasBlass You can also just observe (given $z=a+bi$) that $a^2 + b^2=1$, then it is pretty easy to see that exchanging $a$ for $(1+a)$ and $(1-a)$ will give the desired result. At least that was my first instinct to solve the problem. Jul 27, 2018 at 0:54

Such geometric insights are always valuable, yet it's worth noting that they may not necessarily be unique. For example, the same problem could be solved geometrically using the parallelogram law. Consider that the points $z, -1, -z, 1$ define a parallelogram, then equating the sum of squares of the diagonals with the sum of squares of the sides gives $\,\left(2 |z|\right)^2 + 2^2 = 2 \left(|z-1|^2+|z+1|^2\right)\,$.
(However, the converse does not work, and for proving the parallelogram law from an identity in complex numbers, one would need to start with $\,|a+b|^2+|a-b|^2=2|a|^2+2|b|^2\,$ instead.)