Waste less time on Facebook — follow Brilliant.
×

A query

If \(x\to \infty\) then \(ix\to \infty\) is it true?

where \(i=\sqrt{-1}\)

Note by Tanishq Varshney
2 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

For real \(x\), \(ix\) is purely imaginary and hence it doesn't make sense for it to tend to a value / object which is an element of the extended real number system.

A more accepted / technically correct terminology would be,

\[\lim_{x\to\infty}(ix)=i\infty\]

W|A verification

Prasun Biswas - 2 years, 5 months ago

Log in to reply

What about \( \large \tilde{\infty} \) ? reference .

Guilherme Naziozeno - 2 years, 5 months ago

Log in to reply

Yeah, I had given it much thought earlier. I was also thinking about complex infinity when I first posted that comment. But complex infinity is mostly used to denote a complex value with infinite magnitude but undefined argument.

But in this case, the argument is defined and is \(\dfrac{\pi}2\) (considering the limit is to positive infinity).

W|A verification.

Hence, the term "complex infinity" wouldn't be applicable here, in my opinion.

Prasun Biswas - 2 years, 5 months ago

Log in to reply

hello prasun , i wanted ask u one more thing is \(i^{4 \pi}=1\)

Tanishq Varshney - 2 years, 5 months ago

Log in to reply

Prasun here: Nope, \( i^{4\pi} = (i^4)^\pi = 1^\pi = 1\) is not a valid step as we're dealing with complex numbers so we can't split the exponents as we please. You should do this instead: let \(x = i^{4\pi} \) then \(\ln(x) = 4\pi \ln(i) = 4\pi \cdot \ln(e^{i \pi/2}) = 2\pi^2 i \) or \( x = e^{i \cdot 2\pi^2} = \cos(2\pi^2) + i \sin(2\pi^2)\approx 0.6297 + 0.7769i \).

See this note to spot such pitfalls.

Pi Han Goh - 2 years, 5 months ago

Log in to reply

@Pi Han Goh The main thing that I'd have said (I forgot to reply earlier) is that \((a^b)^c=(a^c)^b=a^{bc}\) doesn't necessarily hold when some of \(a,b,c\) are non-reals.

I had seen a post on MSE regarding this once where an answer explained why the "general rules of exponentiation" doesn't hold when complex numbers are thrown into the mix. I was trying to find that post before I could reply him but it seems you already cleared it up. Thanks!

Also, I'm wondering when did I get to Malaysia?!

Prasun Biswas - 2 years, 5 months ago

Log in to reply

No. Apply de-moivre's theorem you will get to know.

Mahimn Bhatt - 2 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...