A Nice Exponential Equation | Problem 358

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[@XJWill1]

First thing you should do when solving these types of equations is define your functions. The sensible way to do it is to define all functions to be single-value, taking the principal value where necessary. Since z^z is multi-valued in the general case, force it to be single-valued by taking the principal value, which is
z^z = exp(z * Log(z))
where Log(z) is the principal branch of the complex natural Logarithm function
So the equation to solve is
exp( z * Log(z) ) = exp( - pi / 2 )
where I simply inverted the given equation. Now, this could be solved using the complex product-log function, but it is rather tedious to keep track of the conditions for the branch of the product-log and the integer parameter for the solution. Instead, I think it is more informative to solve it graphically. This is best done by splitting the equation into two, one equating the Modulus, and one equating the Arg
exp( x/2 * ln(x^2 + y^2) - y * atan2(y , x) ) = exp( - pi / 2 )
atan2( sin( y/2 * ln(x^2 + y^2) + x * atan2(y, x) ) , cos( y/2 * ln(x^2 + y^2) + x * atan2(y, x) ) ) = 0
To see the solutions on the complex plane, graph those curves and look for the intersection points. If you use desmos note that atan2( y , x ) is entered as arctan( y , x )
There are an infinite number of solutions, but I think the most interesting part of the graph is that there are only two solutions with a negative real part, and they are complex conjugates. In fact, every solution comes with a complex conjugate pair. The single pair with negative real part, +i and -i, and an infinite number of complex-conjugate pairs with positive real part.

[@aplusbi]

Amazing! 😍

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@RyanLewis-Johnson-wq6xs]

z = e^W(-π/2 + 2 i π n), n element Z
z = e^W_(-1)(-π/2 + 2 i π n), Im(W_(-1)(2 i n π - π/2))>-π, n element Z
z = e^W_1(-π/2 + 2 i π n), Im(W_1(2 i n π - π/2))

[@RyanLewis-Johnson-wq6xs]

i^(-i)= e^(π/2) z= ±i

[@scottleung9587]

I got z=-i, but somehow overlooked z=+i.

[@aplusbi]

Let me guess. You replaced -z with something 🤪