Mixing
Baby Rudin 1.14 is essentially Thales's theorem?
Clash Royale CLAN TAG#URR8PPP
up vote
11
down vote
favorite
Exercise 1.14 in Baby Rudin asks to compute
$$
mid 1+zmid^2+mid 1-zmid^2
$$
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 $mathbbC$, so that the image of the circle is the unit circle, and the vertices of the triangle are $-1,z,-z$ for some $zinmathbbC$. 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?
real-analysis geometry complex-numbers soft-question
asked Jul 26 at 23:58
lanskey
544213
add a comment |Â
up vote
11
down vote
favorite
Exercise 1.14 in Baby Rudin asks to compute
$$
mid 1+zmid^2+mid 1-zmid^2
$$
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 $mathbbC$, so that the image of the circle is the unit circle, and the vertices of the triangle are $-1,z,-z$ for some $zinmathbbC$. 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?
real-analysis geometry complex-numbers soft-question
asked Jul 26 at 23:58
lanskey
544213
4
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 $z=costheta+isintheta$ 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.
– Andreas Blass
Jul 27 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.
– lanskey
Jul 27 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.
– Moed Pol Bollo
Jul 27 at 0:54
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Exercise 1.14 in Baby Rudin asks to compute
$$
mid 1+zmid^2+mid 1-zmid^2
$$
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 $mathbbC$, so that the image of the circle is the unit circle, and the vertices of the triangle are $-1,z,-z$ for some $zinmathbbC$. 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?
real-analysis geometry complex-numbers soft-question
asked Jul 26 at 23:58
lanskey
544213
Exercise 1.14 in Baby Rudin asks to compute
$$
mid 1+zmid^2+mid 1-zmid^2
$$
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 $mathbbC$, so that the image of the circle is the unit circle, and the vertices of the triangle are $-1,z,-z$ for some $zinmathbbC$. 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?
real-analysis geometry complex-numbers soft-question
asked Jul 26 at 23:58
lanskey
544213
edited Jul 27 at 0:41
asked Jul 26 at 23:58
lanskey
544213
asked Jul 26 at 23:58
lanskey
544213
asked Jul 26 at 23:58
lanskey
544213
544213
4
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 $z=costheta+isintheta$ 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.
– Andreas Blass
Jul 27 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.
– lanskey
Jul 27 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.
– Moed Pol Bollo
Jul 27 at 0:54
add a comment |Â
4
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 $z=costheta+isintheta$ 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.
– Andreas Blass
Jul 27 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.
– lanskey
Jul 27 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.
– Moed Pol Bollo
Jul 27 at 0:54
4
4
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 $z=costheta+isintheta$ 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.
– Andreas Blass
Jul 27 at 0:11
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 $z=costheta+isintheta$ 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.
– Andreas Blass
Jul 27 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.
– lanskey
Jul 27 at 0:17
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.
– lanskey
Jul 27 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.
– Moed Pol Bollo
Jul 27 at 0:54
@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.
– Moed Pol Bollo
Jul 27 at 0:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
(Too long for a comment.)
It is a valid proof of Thales' theorem, which I believe has been re-discovered many times.
In the context of Baby Rudin's book, I don't expect that it was meant to be "found" as such.
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 $,left(2 |z|right)^2 + 2^2 = 2 left(|z-1|^2+|z+1|^2right),$.
(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.)
answered Jul 27 at 0:54
dxiv
53.8k64796
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
(Too long for a comment.)
It is a valid proof of Thales' theorem, which I believe has been re-discovered many times.
In the context of Baby Rudin's book, I don't expect that it was meant to be "found" as such.
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 $,left(2 |z|right)^2 + 2^2 = 2 left(|z-1|^2+|z+1|^2right),$.
(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.)
answered Jul 27 at 0:54
dxiv
53.8k64796
add a comment |Â
up vote
2
down vote
(Too long for a comment.)
It is a valid proof of Thales' theorem, which I believe has been re-discovered many times.
In the context of Baby Rudin's book, I don't expect that it was meant to be "found" as such.
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 $,left(2 |z|right)^2 + 2^2 = 2 left(|z-1|^2+|z+1|^2right),$.
(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.)
answered Jul 27 at 0:54
dxiv
53.8k64796
add a comment |Â
up vote
2
down vote
up vote
2
down vote
(Too long for a comment.)
It is a valid proof of Thales' theorem, which I believe has been re-discovered many times.
In the context of Baby Rudin's book, I don't expect that it was meant to be "found" as such.
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 $,left(2 |z|right)^2 + 2^2 = 2 left(|z-1|^2+|z+1|^2right),$.
(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.)
answered Jul 27 at 0:54
dxiv
53.8k64796
(Too long for a comment.)
It is a valid proof of Thales' theorem, which I believe has been re-discovered many times.
In the context of Baby Rudin's book, I don't expect that it was meant to be "found" as such.
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 $,left(2 |z|right)^2 + 2^2 = 2 left(|z-1|^2+|z+1|^2right),$.
(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.)
answered Jul 27 at 0:54
dxiv
53.8k64796
answered Jul 27 at 0:54
dxiv
53.8k64796
answered Jul 27 at 0:54
dxiv
53.8k64796
answered Jul 27 at 0:54
dxiv
53.8k64796
53.8k64796
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863936%2fbaby-rudin-1-14-is-essentially-thaless-theorem%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
4
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 $z=costheta+isintheta$ 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.
– Andreas Blass
Jul 27 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.
– lanskey
Jul 27 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.
– Moed Pol Bollo
Jul 27 at 0:54