Is $\pi$ a complex number?
I just came across an exam paper the answer of one of the questions about complex numbers says $\pi$ is a complex number. How $\pi$ is represented as a complex number?
$\endgroup$ 45 Answers
$\begingroup$Every real number is a complex number. Therefore $\pi$, which is a real number, is a complex number.
$\pi$ is not an imaginary number, which are numbers in the form of $xi$, $x \in \mathbb R$.
$\endgroup$ 2 $\begingroup$Since $\mathbb R\subseteq \mathbb C$ and $\pi\in\mathbb R$, it follows that $\pi \in\mathbb C$
$\endgroup$ 2 $\begingroup$a complex number is an ordered pair of real numbers ($x = (a, b)$). when we say $\pi$ is a complex number, we simply mean $(\pi, 0)$.
$\endgroup$ $\begingroup$π is a real number. real numbers are subset of complex numbers. so every element of real numbers is in complex numbers for this reason π is a complex number and show it as a ordered pair (π , 0) or π+i0
$\endgroup$ 2 $\begingroup$Yes. $\pi$ is a complex number.
It is known that $\pi\in\mathbb{R}$, $\mathbb{R}\subset\mathbb{C}$. Therefore, $\pi\in\mathbb{C}$.
$\endgroup$
