0

Plotting the function on the real axis reveals that it has three solutions

enter image description here

Therefore the solution set is:

$$ z = \left\{-\frac{2}{\ln2}W\left(\frac{\ln2}{2}\right),2,4\right\} $$

But plotting in the complex plane reveals that there are other solutions far away from the real axis.

enter image description here

The numerical approximation result is:

$$ z≈7.65±11.95i $$

How should we give the expression for the complex solution of the original equation?

Aster
  • 1,220
  • wolfram gives answer using W functkion. https://www.wolframalpha.com/input?i=solve+2%5Ez+%3D+z%5E2 – coffeemath Jan 26 '24 at 04:06
  • To see the pattern of roots, I recommend plotting the equations Re(2^z - z^2) = 0 and Im(2^z - z^2) = 0. The complex roots will be those points common to both loci. (Where 2^z denotes exp(ln(2)*z).) – Dan Asimov Jan 26 '24 at 05:41

1 Answers1

2

The Lambert $W$ function is multivalued, with infinitely many branches. The one you're taking as $W(\ln(2)/2)$ is the "$0$" branch. The complex roots correspond to the other branches. Branches $1$ to $5$ correspond to complex roots approximately

$$ 7.654506496- 11.95367891 \,\mathrm{i}, 10.03684857- 30.81806330 \,\mathrm{i}, 11.31574563- 49.20372800 \,\mathrm{i}, 12.19872547- 67.46928744 \,\mathrm{i}, 12.87413263- 85.68453064 \,\mathrm{i} $$

and $-1$ to $-5$ the complex conjugates of those.

Robert Israel
  • 448,999