Does e^z - z^2 = 0 Have Infinite Solutions?

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SUMMARY

The equation ez - z2 = 0 has infinitely many solutions due to the periodic nature of the exponential function ez. By rewriting the equation as z2 = e2 log z, we derive the form e{z - 2 log z} = 1. This leads to the conclusion that the solutions can be expressed as z - 2 log z = 2πn, where n is an integer. The Lambert W function is identified as a useful tool for solving this equation.

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Likemath2014
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How can we show that the following equation has infinitely many solutions
e^z-z^2=0.
Thanks
 
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Work within a fixed branch of logz and write : ## z^2=e^{2logz} ##

Once you find a solution, you have infinitely-many, by periodicity of ##e^z##.
 
Continuing WWGD's post: you get ##e^z=e^{2\log z}## and thus the equation ##e^{z-2log z}=1##. In other words you are looking at the solutions of each of the equations ##z-2\log z=2\pi n## for ##n\in\mathbb{Z}##.
 
It seems that something like Lambert's W function may be helpful in finding solutions to platetheduke's equation.
 

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