Practice Problem for Advanced Topics

$\begin{aligned} e^{i\pi} +(\log 1) ^2 & =x-0!\\ \cos \pi +i\sin \pi +0^2&=x-1 \\ - 1+i(0)+0&=x-1 \\ - 1&=x-1\\ \Rightarrow x&=\boxed{0} \end{aligned}$

e^ipi+1=0 famous expression

$\large e^{i*\pi}=Cos \pi + iSin \pi=-1+i*0=-1.$

This problem requires knowing some facts about the properties of these numbers. For example, $0! = 1$ , $e^{i\pi} = -1$ , and $\log 1 = 0$ . $0^{2} = 1$ so $-1 + 0 + 1 = 0$ and $x = 0$ .