Jeez! It's impossible?

$$ Note that: $$ $i^2=-1$ $$ and $$ $e^{\frac{2n-1}{2}i\pi}=\cos\left(\frac{2n-1}{2}i\pi\right)+i\sin\left(\frac{2n-1}{2}i\pi\right)=i\quad\text{for }n\text{ is odd}.$ $$ Therefore $$ $\begin{aligned} \ln(-1)&=\ln i^2\\ &=\ln\left(e^{\frac{2n-1}{2}i\pi}\right)^2\\ &=\ln\left(e^{(2n-1)i\pi}\right)\\ &=(2n-1)i\pi. \end{aligned}$ $$ Hence, $p^2=((2n-1)i)^2=-(2n-1)^2$ and $\max\left[p^2\right]$ is obtained when $n=1$ . Thus, $\max\left[p^2\right]=-(2(1)-1)^2=\boxed{\color{#3D99F6}{-1}}$ . $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

We can solve it in an easier way... Note that taking e^ taking from both sides' we get e^pπ=-1.... Noiw famous Euler's eq, e^ix= coax + isinx,, and comparing, we get p=I= sqrt(-1)....$==> p^2= -1...

Given that $\ln{(-1)} = p \pi\quad \Rightarrow e^{p\pi} = -1$

This implies that $p = i(2n+1)$ , where $n = 0, 1, 2, ...$ and $i = \sqrt{-1}$ . This is because $e^{i(2n+1)\pi} = \cos {(2n+1)\pi} + i \sin {(2n+1)\pi} = -1$

Since $p^2 = i^2(2n+1) = -(2n+1)$ , this means that $min[p^2] = \boxed {-1}$ , when $n=0$ .

as ln(-1)= 3.14i so just sustitute this value in the given equation we will get the answer