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A query

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

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

Note by Tanishq Varshney
1 year, 6 months ago

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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 · 1 year, 6 months ago

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@Prasun Biswas What about \( \large \tilde{\infty} \) ? reference . Guilherme Naziozeno · 1 year, 6 months ago

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@Guilherme Naziozeno 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 · 1 year, 6 months ago

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@Prasun Biswas hello prasun , i wanted ask u one more thing is \(i^{4 \pi}=1\) Tanishq Varshney · 1 year, 6 months ago

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@Tanishq Varshney 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 · 1 year, 6 months ago

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@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 · 1 year, 6 months ago

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@Tanishq Varshney No. Apply de-moivre's theorem you will get to know. Mahimn Bhatt · 1 year, 6 months ago

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