# Is $\infty+ i\times \infty$: $\tilde{\infty}$?

Mathematica would leave $\infty+i*\infty$ there, no more simplification.

Before I ask the question, there are number of assumptions I have. If any of them is not right, please point it out. Thank you!

1. I assume what Mathematica means when displaying $i*\infty$ is $\infty$ in imaginary positive direction. Because $(1+i)*\infty$ would be simplified into $\frac{1+i}{\sqrt{2}}\infty$. The first part is simplified into unit length. Besides, there is only one infinity in Riemann sphere.

2. If assumption 1 is right: I also think $\infty+i*\infty$ evaluate into $\infty$ which can have any direction between real positive direction and imaginary positive direction inclusive, and that's why Mathematica leave the expression alone.

If these two assumptions are both right, here comes my question:

If the angle between directions of lhs and rhs infinity in an addition is $180~^{\circ}$ (or $2\pi$), the result is indeterminate, or undefined. Except this, is $\infty$ with a range of direction complex infinity ($\tilde{\infty}$)?

Thank you again!

• The complex positive infinity would just be an infinitely long vector in the $(1,1)$ direction – Anthony P Dec 5 '16 at 5:42
• @AnthonyP So Mathematica treats infinity as complex positive infinity, not the infinity in Reimann sphere? – Zhigang An Dec 5 '16 at 5:47
• I'm voting to close this question as off-topic because it belongs to mathematica.se. – Did Dec 5 '16 at 7:36
• @Did Though I mentioned Mathematica in the problem, the real question is not relevant to Mathematica, or any Maths tool software. My question is "Is infinity + i*infinity = complex infinity". – Zhigang An Dec 5 '16 at 7:39
• Reading very charitably your question, it seems you might be asking whether, if $(z_n)$ and $(w_n)$ are complex sequences such that $$|z_n|\to\infty\quad |w_n|\to\infty\quad\lim\arg z_n=\alpha\quad\lim\arg w_n=\beta$$ with $$\alpha\ne\beta$$ then... then what? Then, do we have $$\lim|z_n+w_n|=+\infty\ ?$$ Is this your question? Then 1. it does not show in the current text and 2. the answer is obvious, no? If you insist that the question is not about Mathematica (which frankly I fail to understand, see for example your first comment), one could close this as "unclear". – Did Dec 5 '16 at 7:49

Certainly, if $(a_n)$ and $(b_n)$ are real sequences tending to $\infty$ at (essentially) the same rate, then $\lim \arg (a_n+b_ni)=\pi/4$, a known quantity. Alternatively, if $(a_n)$ and $(b_n)$ are real sequences tending to $\infty$ at unspecified rates, then even though $\lim \arg (a_n+b_ni)$ may not exist, the arguments (and potential limit value) of $a_n+b_ni$ stay in $[0,\pi/2]$.
If either case captures what you meant by $\infty+i\infty$, then the answer may be "it's not Complex Infinity because you know at least some bounds on the arguments; the limiting behavior of the argument is not completely unknown or undefined."
That may or may not explain why Mathematica does not turn $\infty+i\infty$ into $\widetilde{\infty}$. That would be a question about Mathematica.