@@HaniSantosa In physics, I often wonder where complex numbers are used only for ease, and where we can’t do without them at all because nature is such. Is there a theorem that can show that every calculation or derivation that can be done with complex numbers can also be done without them?
@@rohitrathi4552 you always can do things without them by converting to Vectors and rotation matrixes. That's because they're equivalent to each other.
Since the rotation is 90 deg the rotation matrices are trivial, and the result will to just flip the x and y and negate one of them (which one depends on clockwise vs anticlockwise). The results along the way will look very similar to the complex plane version shown here, and I don’t feel like it was any harder, maybe even slightly easier, especially if you are not all that familiar with complex numbers vs vectors.
There is a similar puzzle in George Gameau‘s 1,2,3 …Infinity. I remember reading that many many years ago. You brought back a wonderful memory. And of course I learned something new today as well. Thank you.
One can solve the problem purely geometrically: let the Shed be at S, Pine and Willow at P and W while their respective markings at P' and W'. Moreover let W'' and P'' be such that PWP''W'' is a clockwise square. Consider the clockwise 90 degrees rotation around W composed with the clockwise 90 degrees rotation around P. The transformation in total rotates the plane about 180 degrees and thus must be a point symmetry around some point X. Since the first transformation moves W' to S and the second moves S to P' their composition moves W' to P' and thus X is our point of interest - midpoint of P'W'. But the first transformation leaves W intact while the second moves it to W'' so that W moves to W'' in the composition. So X must also be the midpoint of WW'' i.e. the center of the square. But the square can be constructed just based on the points P and W alone which is what we were after.
There are some people saying "hey we can also do this by useing geometry" but on of the points of complex numbers is doing geometry with an easier to manipulate language.
Quickly scribbling some notes in my notebook: p, w, s are points. p' = p + i(p - s), w' = w - i(w - s), x = ½(p' + w') = ½(p + i(p - s) + w - i(w - s)) = ½(p + ip - is + w - iw + is) = ½(p + w + i(p - w)) Oh look, there's no s in the result, regardless of p, w, or s.
Hi,my dear companion Brian,I hope you're brilliant!One of the biggest questions occupying my mind for so long is why we've invented imaginary numbers providing we know that they're not for real!Thank you for actioning my question!😊😊 I wish you the best and break a leg!
“Imaginary” numbers are no more real than reals, or even natural numbers. Gauss even proposed to call i a lateral unit, but that didn’t seem to take root. Even now that we know there are several nice 2-dimensional real algebras-not just complex numbers-it’d still be insightful.
you can solve it in a cartesian way. the Shed is in (a,b) Pine (1,0) and Willow (-1,0) The positive point will be (1 + b , 1 - a) The negative point will be ( -1 - b , 1 + a) to get half of the distance you do (x1 - x2)/2 doing that: ((1+b) - (-1-b))/2 = 1 + b Get the negative point and sum the half distance (-1 - b) + (1 + b) = 0 The X of the treasure will be 0 doing the same with the y: ((1 - a) - (1 + a))/2 = -a (1 + a) + (-a) = 1 The Y of the treasur will be in 1 Tresure (0,1)
You can solve the puzzle this way : Maybe the position of the shed does not matter, it is just a trap : Imagine the shed is somepoint. Do the calculations and find the treasure point. Then imagine again the shed is some other point and do the calculations. You'll find the same treasure point. To be sure choose again another place for the shed. The treasure point will be again the same. You can conclude that the position of the shed does not matter, and you have found your treasure point :)
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Could you have solved this problem without using complex numbers but using only the Cartesian coordinate system and vectors?
good idea
I think it's doable, but it becomes more complicated (involving rotation matrix and so on)
@@HaniSantosa In physics, I often wonder where complex numbers are used only for ease, and where we can’t do without them at all because nature is such. Is there a theorem that can show that every calculation or derivation that can be done with complex numbers can also be done without them?
@@rohitrathi4552 you always can do things without them by converting to Vectors and rotation matrixes. That's because they're equivalent to each other.
Since the rotation is 90 deg the rotation matrices are trivial, and the result will to just flip the x and y and negate one of them (which one depends on clockwise vs anticlockwise). The results along the way will look very similar to the complex plane version shown here, and I don’t feel like it was any harder, maybe even slightly easier, especially if you are not all that familiar with complex numbers vs vectors.
Instructions unclear: ended up getting detention and $2,345,012.83 from the federal government.
Imagine whoever wrote that in the textbook was just joking
Fermat
@@alejrandom6592 but he wouldn't have enough space for the treasure
You say imagine, but maybe they were joking for real.
thats true, its in the imaginary axis, so it isnt real
There is a similar puzzle in George Gameau‘s 1,2,3 …Infinity. I remember reading that many many years ago. You brought back a wonderful memory. And of course I learned something new today as well. Thank you.
One can solve the problem purely geometrically: let the Shed be at S, Pine and Willow at P and W while their respective markings at P' and W'. Moreover let W'' and P'' be such that PWP''W'' is a clockwise square. Consider the clockwise 90 degrees rotation around W composed with the clockwise 90 degrees rotation around P. The transformation in total rotates the plane about 180 degrees and thus must be a point symmetry around some point X. Since the first transformation moves W' to S and the second moves S to P' their composition moves W' to P' and thus X is our point of interest - midpoint of P'W'. But the first transformation leaves W intact while the second moves it to W'' so that W moves to W'' in the composition. So X must also be the midpoint of WW'' i.e. the center of the square. But the square can be constructed just based on the points P and W alone which is what we were after.
Damn
Does that mean the treasure is also imaginary or simply one unit above the midpoint of the trees?
the true treasure is the vectors and complex numbers we made along the way
@@eterty8335 hahahaha
The treasure! The treasure is imaginary!!
The treasure wasn't there, it was only imaginary.
There are some people saying "hey we can also do this by useing geometry" but on of the points of complex numbers is doing geometry with an easier to manipulate language.
This is a super awesome way to story tell around the complex plane. Great video, Brian! 🎉
hypothetically would be cool if there was some kind of class assignment as a physical analogue to this problem.
Do you know where the treasure is?
Aye.
Where's the treasure?
i.
No, seriously, where's the treasure?
.
.
.
;)
This was a great video, thank you Brian!!
The real treasure was the math we learned on the way
That is one way for complex numbers professor to make student interested... Shed was always imaginary to begin with!
But How Can we sure that the Distance Between Both tree is 2unit??
That's just a metric. If the distance were, 4 for example, then the treasure would be at 2i
You can always rescale accordingly.
Thankyou Dear
Can we say Treasure lies on Perpendicular Bisector of Both tree.
What teachers think the real world will be like:
Great video as always!!!!
Thanks for the wonderful video Brian! (I think that was your name... or was it?!)
Can we find the position of the shed via this method as well?
Nope. We aren't given any information that'd help us figure that out.
You can just imagine the shed at some point and hope its position doesn't matter - that's true, but you don't need to prove that
Now just count i steps from the mid point of the trees.
Can't you just test multiple values for a+bi and follow the directions from there?
Awesome! The math tells me I just have to walk "i" paces!
"Wait, fu-"
Quickly scribbling some notes in my notebook: p, w, s are points. p' = p + i(p - s), w' = w - i(w - s), x = ½(p' + w') = ½(p + i(p - s) + w - i(w - s)) = ½(p + ip - is + w - iw + is) = ½(p + w + i(p - w)) Oh look, there's no s in the result, regardless of p, w, or s.
My solution assumed the shed is at 0 and the trees are at zP and zW
Wait, why are both the pine and willow both a+bi? Shouldn’t one be c+di?
The pine and willow are at 1 and -1, it's the shed that's at a+bi.
You just said it was "impossible"....then you proceed and solve it yourself....
Its a=x b=if you imagine the x axis you get a y axis.
Idk.
No, I cannot! ...and I'm ok with that 😊
Bro we can use geometry
You didn't even tell us what the treasure was. Very disappointed.
I could not get it Weak Student I guess but there is an urge inside me which wanna know this stuff.
Hi,my dear companion Brian,I hope you're brilliant!One of the biggest questions occupying my mind for so long is why we've invented imaginary numbers providing we know that they're not for real!Thank you for actioning my question!😊😊 I wish you the best and break a leg!
You should watch veritasium’s video on it, it’s really well made
“Imaginary” numbers are no more real than reals, or even natural numbers. Gauss even proposed to call i a lateral unit, but that didn’t seem to take root. Even now that we know there are several nice 2-dimensional real algebras-not just complex numbers-it’d still be insightful.
So the treasure was imaginary 💀
you can solve it in a cartesian way.
the Shed is in (a,b)
Pine (1,0) and Willow (-1,0)
The positive point will be (1 + b , 1 - a)
The negative point will be ( -1 - b , 1 + a)
to get half of the distance you do (x1 - x2)/2
doing that:
((1+b) - (-1-b))/2 = 1 + b
Get the negative point and sum the half distance
(-1 - b) + (1 + b) = 0
The X of the treasure will be 0
doing the same with the y:
((1 - a) - (1 + a))/2 = -a
(1 + a) + (-a) = 1
The Y of the treasur will be in 1
Tresure (0,1)
why did you take the coordinates of the trees as (1,0) and (-1,0) and not something else like (2,0) and (-3,0). What's the reason?
Ahahaha...
No.
You can solve the puzzle this way : Maybe the position of the shed does not matter, it is just a trap :
Imagine the shed is somepoint. Do the calculations and find the treasure point.
Then imagine again the shed is some other point and do the calculations. You'll find the same treasure point. To be sure choose again another place for the shed. The treasure point will be again the same.
You can conclude that the position of the shed does not matter, and you have found your treasure point :)
I'm a 10th grade student and I love maths & thanks for making such cool and informative videos.
Love from KASHMIR❤
Small world....Also from Kashmir.
@@MohdIrfanZ7 A/a brother
@@NINJA-tf6bf Walaikum asalam dear.
Yaaaayyy!
Do you also talk like this in a conversation with a friend? Just wondering.
treasure
1 + i (√ −x + 0.1) = 0.9999