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

[Likemath2014]

How can we show that the following equation has infinitely many solutions
$$e^z-z^2=0$$.
Thanks

 

[WWGD]

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 $e^z$.

 

[platetheduke]

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 $n\in\mathbb{Z}$.

 

[WWGD]

It seems that something like Lambert's W function may be helpful in finding solutions to platetheduke's equation.

 

[blue_raver22]

for your question. The equation e^z-z^2=0 is a complex exponential equation, where z is a complex number. In order to show that this equation has infinitely many solutions, we can use the fundamental theorem of algebra, which states that a non-constant polynomial equation has exactly n complex solutions, where n is the degree of the polynomial. In this case, the degree of the polynomial is 2, so the equation has exactly 2 complex solutions. However, since z is a complex number, it can take on infinitely many values. Therefore, there are infinitely many solutions to this equation, as every complex number can be written in the form z=a+bi, where a and b are real numbers. This means that for every value of z, there exists a solution to the equation e^z-z^2=0. Additionally, we can also graph the equation in the complex plane and see that it forms a parabola, which will intersect the x-axis (where the value of z is 0) an infinite number of times, further supporting the fact that there are infinitely many solutions to this equation.