This is why a length of a vector on a complex plane is |z|=√(z*×z). z* is a complex conjugate of z.
I’ve noticed that, if an equation calls for a number squared, they usually really mean a number multiplied by its complex conjugate.
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You need to add some disclaimer to this diagram like “not to scale”…
It’s to scale.
Which scale is left as an exercise to the reader.
I really don’t think it is.
I may not have been entirely serious
Isn’t the squaring actually multiplication by the complex conjugate when working in the complex plane? i.e., √((1 - 0 i) (1 + 0 i) + (0 - i) (0 + i)) = √(1 + - i2) = √(1 + 1) = √2. I could be totally off base here and could be confusing with something else…
I think you’re thinking of taking the absolute value squared, |z|^2 = z z*
Now calculate the angles
That’s actually pretty easy. With CB being 0, C and B are the same point. Angle A, then, is 0, and the other two angles are undefined.
A is clearly a right angle
A is drawn in such a way that it resembles a right angle, but it is not labeled as such. The length of the hypotenuse is given as zero. The opposite angle cannot be anything but 0°.
The pythagoras theorem only holds if A is a right triangle
What is depicted here isn’t even a polygon, let alone a triangle, let alone a right triangle. This is just a line segment. Line AB is the same as line AC. There is no line BC. BC is a single point.
I suppose it could possibly depict a weird cross section of two orthogonal circles in a real and an imaginary plane.
No thank you
Too complexe for me ;)
Doesn’t this also imply that
i == 1becauseCBhas zero length, forcingACandABto be coincident? That sounds like a disproving contradiction to me.I think BAC is supposed to be defined as a right-angle, so that AB²+AC²=CB²
=> AB+1²=0²
=> AB = √-1
=> AB = i
I mean, I see that’s how they would have had to get to i, but it’s not a right triangle.
The length would be equal to the absolute value
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Turn around…
Bright eyes.
Every now and then, do ya fall apart?
Turn around…
Exactly what I thought of, but then I was like… nah that’s too cheesy
Too cheesy?!
Every now and then, I get a little bit lonely and you’re never coming 'round
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What if not a Hilbert space?
Looks like a finite state machine or some other graph to me, which just happens to have no directed edges.
