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- เผยแพร่เมื่อ 14 ต.ค. 2024
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I really wish you had plotted the solutions for us.
See my other comment for the equations you can plot to visualize the infinite number of solutions.
Which app do you use?
Notability
To visualize in the complex plane the infinite number of complex z values that solve the equation,
exp( Log(z) / z ) = i
plot the following sets of curves and observe the intersection points
0 = x * ln(x^2 + y^2) + 2*y * atan2(y , x)
1 = tan( (1/2 *x*atan2(y , x) - 1/4 *y*ln(x^2 + y^2) ) / (x^2 + y^2) )
If using Desmos, note that atan2(y , x) is entered as arctan(y , x)
The curves become difficult to resolve as you zoom in, like fractals. Desmos sometimes creates artifacts in the graphs, but if you keep zooming in you can get better resolution.
One thing that is interesting to see from the graph, that I think is difficult to realize if the solution is given in terms of the product-log function, is that there is only 1 solution with negative real part (real part is approximately - 1.86). Any of the other infinite number of solutions have positive real part.
I tried it and got a very different wrong answer.... 😅 but still, great problem and great solution!
Thanks!
Cool!
U are fun❤😄
u is funner than me 😜
Not as straightforward as it first appears. I rearranged to get
u = 1/z
u^u = -i
Which didn't simplify!
A nice way to approach it, though
@mcwulf25
It appears that you assumed (1/u)^u = 1 / u^u
I wonder if you proved that to be the case. Because in general
(1 / w)^p is not equal to 1 / w^p
For example, w = - 1 and p = 1/2 demonstrates the inequality of the more general case
For (1/u)^u and 1 / u^u , try u = - 1/2
It does actually simplify!! There is this super cool function called the "super square root," which is defined such that ssrt(x) is the inverse of x^x (it can also be calculated through ssrt(x) = e^(W(lnx)).) Therefore, you can say:
u^u = -i
u = ssrt( -i )
1/z = ssrt( -i )
z = 1/(ssrt( -i ))
= 1/ (e^(W(ln(-i))))
= e^-(W(-i*pi/2)) (which is the same as the the answer he gave! - it is just the 'principal value' or where n = 0)
≈ 0.438 + 0.361i
Your mistake is equating ln( z ^ (-1) ) = -1 * ln(z)
That is not true in general. It is only an identity for real z > 0
For other values of z, equality does not always hold
Can you point out a particular value of z which does not hold for ln( z ^ (-1) ) = -ln(z)?
@@xleph2525 The most obvious failure is - 1
How is the problem solved without making this mistake?
@@XJWill1 I see your point, but nothing like -1 was being operated on in the string of reasoning. In fact, it is very apparent that no real z < 0 can be a solution. I suppose there could have been a disclaimer in the video with that observation, but I don't believe that would have been necessary.
On a side note, I think -ln(z) = ln( z^-1 ) does hold true for z = a + bi. I encourage you to try a few complex numbers and see what happens.
@@xleph2525 You obviously do not understand identities and real and complex numbers.
Ну и чему же равно z?
Okay, but I actually do prefer tea. 😅
Absolutely: |Tea| 😁