Is $\pi$ a complex number?

Question

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?

The set of real numbers is a subset of the set of complex numbers. — Extremal, Oct 16 '15 at 17:55
Related: https://math.stackexchange.com/questions/1409398/can-a-real-number-be-added-to-a-complex-number/1409404#1409404 — Patrick Stevens, Oct 16 '15 at 17:56
"Bye_World"'s comment is in some ways better than the two answers posted so far. ${}\qquad{}$ — Michael Hardy, Oct 16 '15 at 18:06

score 6 · Answer 1 · answered Oct 16 '15 at 17:55

6

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 $xi$, $x \in \mathbb R$.

answered Oct 16 '15 at 17:55

wythagoras

25,026

i thought the same too thank!! – user256670 Oct 16 '15 at 19:37
5

@user256670 The highest form of thanks on math.SE is an upvote and accept. Any answers that you found useful you should upvote by clicking the up arrow left of the answer (it'll turn red once you do) -- you can upvote multiple answers to a single question. And for the answer that helped you the most you should accept by clicking the checkmark left of the answer (it'll turn green once you do) -- you can only accept a single answer to a given question. – Oct 16 '15 at 20:44

score 2 · Answer 2 · answered Oct 16 '15 at 17:55

2

Since $\mathbb R\subseteq \mathbb C$ and $\pi\in\mathbb R$, it follows that $\pi \in\mathbb C$

answered Oct 16 '15 at 17:55

5xum

123,496
6
128
204

Even more, $\mathbb R\subset \mathbb C$. – marty cohen Oct 16 '15 at 18:53
agree!!! thanks!! – user256670 Oct 16 '15 at 19:37

score 0 · Answer 3 · answered Nov 08 '17 at 20:03

0

a complex number is an ordered pair of real numbers ($x = (a, b)$). when we say $\pi$ is a complex number, we simply mean $(\pi, 0)$.

answered Nov 08 '17 at 20:03

guanglei

793

mathletic · Answer 4 · 2018-06-10T05:08:38.157

0

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

edited Jun 10 '18 at 05:08

answered Jun 02 '18 at 13:31

mathletic

69

1

Welcome to MSE. It is in your best interest that you use MathJax. – José Carlos Santos Jun 02 '18 at 13:39
You are responding to a Question from two years ago. It is welcome to post a new Answer if you have some new information to add, but merely repeating observations that were made at the time the Question was posted does not add useful content. – hardmath Jun 02 '18 at 13:55

score 0 · Answer 5 · answered May 15 '19 at 14:38

0

Yes. $\pi$ is a complex number.

It is known that $\pi\in\mathbb{R}$, $\mathbb{R}\subset\mathbb{C}$. Therefore, $\pi\in\mathbb{C}$.

answered May 15 '19 at 14:38

Bernard

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Is $\pi$ a complex number?

5 Answers5