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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
2 years, 2 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 · 2 years, 2 months ago

What about $$\large \tilde{\infty}$$ ? reference . · 2 years, 2 months ago

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).

Hence, the term "complex infinity" wouldn't be applicable here, in my opinion. · 2 years, 2 months ago

hello prasun , i wanted ask u one more thing is $$i^{4 \pi}=1$$ · 2 years, 2 months ago

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. · 2 years, 2 months ago

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?! · 2 years, 2 months ago