Solving eⁱᶻ = iz | Problem 369

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[@jimschneider799]

The power-tower solution you gave at the end of the video is ill defined, even though every step along the way can be rigorously justified. However, you cannot take the limit, because i*z is not equal to e. The best you can do is an arbitrary, but finite, power tower, where the innermost exponent is i*z.

[@XJWill1]

There are an infinite number of solutions. To visualize them in the complex plane, graph the following implicitly defined curves and look for the intersection points: exp( -2 * y ) = x^2 + y^2 atan2( sin(x) , cos(x) ) = atan2( x , - y ) If you are using desmos, atan2( y , x) is entered as arctan( y , x ) All of the solutions have negative imaginary part. The real part of the solutions is an even function in the sense that the solutions come in pairs (x , y) and (-x , y)

[@朕是神]

I don't think the last minute of your video makes sense when talking about a number that blows up like that. IIRC it only works when the series converge, which in general e^e^e^..... does not.

[@rajeevlochan5499]

I have a question ,hope you will solve it If x^2 + x=1 then the value x^7+34/x+2 Please solve it ,I would be obliged

[@mihir_666]

W(-1) = -0.31813 + 1.33724i While e^(infinite e) is infinity So your analysis is wrong.

[@mox3909]

Yeah, just because there are an infinite stack of e's in the exponent, there's still a variable at the top. Both f(x)=x and g(x)=e^x extend to infinity, but it's pretty bold to claim that f(inf)=g(inf)...

[@RexxSchneider]

exp(iz) = iz --> exp(-iz) = 1/iz --> -iz.exp(-iz) = -1 --> -iz = W(-1) --> z = iW(-1) That's four (or at most five) steps, and it doesn't take 10 minutes. Where you have a mix of polynomial and exponent unknowns, you know it almost always requires the use of the Lambert W-function, so you start from that. The approximate answer is z = - 1.33724 - 0.31813i for the first value of W(-1).

[@aplusbi]

Nice!

[@scottleung9587]

Very cool!

[@aplusbi]

Thanks

[@trojanleo123]

z = i*W(-1)

[@mcwulf25]

-iz = W(-1)

[@jmlfa]

I hate Lambert !

[@rajeevlochan5499]

Hey bro