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CTK Exchange
alexb
Charter Member
2758 posts |
Sep-18-10, 10:09 PM (EST) |
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1. "RE: A complex number approach..."
In response to message #0
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>In a recent discussion during a high school math class, my >students learned that a + bi multiplied by it's conjugate >gives the real number a^2 + b^2. After reading >https://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/ >(which I found interesting) I thought how the number a + bi >being rotated by a - bi winds up being a number on the >positive real axis, which has the same length as either >conjugate. > >I think this is really cool and would make for a great >animation. Not sure if this might qualify as a proof. You mean "... of the Pythagorean theorem" of course. But what is exactly a proof here? >I >have a feeling this might turn into another "How do you know >that c is the radius of your circle?" No, much simpler. What is the length of a complex number? And, perhaps, where did it come from? |
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jmolokach
Member since Jan-11-11
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Sep-23-10, 02:02 PM (EST) |
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4. "RE: A complex number approach..."
In response to message #3
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Well in using the last diagram I seem to have 2 similar right triangles. If we call the original blue length c, then c/a = (a^2 + b^2)/c which implies a^2 + b^2 = c^2 / a ??? Why is there an extra a? Either the complex numbers makes this problem messed up or my triangles aren't similar or I am just not seeing the error. I am siding with the latter... Can anyone explain this? molokach |
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alexb
Charter Member
2758 posts |
Sep-23-10, 10:59 PM (EST) |
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11. "RE: A complex number approach..."
In response to message #10
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>This is really cool. I think I will make a poster out of >this and put on my classroom wall... I wonder if this might >go under the PWW category? Of course. >It'seems like one could do this >for rational numbers as well as integers... just make the >distance between the dots 1/n, where n is a common >denominator.... Yes. Rational and integers are all the same around here. >I am still stuck on why that triangle in the complex numbers >diagram has no geometric relationship to the right >triangle... It seems like there should be a way to show >that the length along the real axis is the square of the >length of the vector from a + bi without using the PT. Have >you heard of anyone doing this? No, I haven't.
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jmolokach
Member since Jan-11-11
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Sep-24-10, 06:17 AM (EST) |
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13. "RE: A complex number approach..."
In response to message #11
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the aforementioned site mentons as part of the proof: "Multiplying a vector a + i b with angle A and length r by its complex conjugate a - ib gives a complex number with angle 0 = A + (-A) and length r2 units according to the add the angles, multiply the lengths polar coordinate, multiplication rule..." is this "rule" based in the PT? if not then this has major ramifications...even to the point of validating my calculus proof... This is why I mentioned originally "Is this another 'How do you know that c is the radius of your circle?'" It's just that this time it's r instead of c... molokach |
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alexb
Charter Member
2758 posts |
Sep-24-10, 06:23 AM (EST) |
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15. "RE: A complex number approach..."
In response to message #13
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>is this "rule" based in the PT? This rule is based on the definition of multiplication of complex numbers. >if not then this has major ramifications...even to the point >of validating my calculus proof... You may try discussing this on a larger forum, say, at one of mathforum.org groups. In my view, as long as the definition of length requires PT, whatever comes afterwards can't be used to prove the theorem. |
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