Limits involving logarithm and argument in the complex plane

Question

• $\operatorname{Log}((2/n) + 2i)$ as $n \to \infty$
• $\operatorname{Log}(2 + (2i/n))$ as $n \to \infty$
• $\operatorname{Arg}((1+i)/n)$ as $n \to \infty$
• $(\operatorname{Arg}(1+i))/(n)$ as $n \to \infty$

For the Log questions, I am getting $(i\pi)/2 + \log(2)$ for the first problem, then for the second I am getting only $\log(2)$. Because the Log's in the questions are capitalized, I think I may have to add on $2\pi i k$ to each of the answers. Is that correct?

for the last two problems (the Arg problems) I got zero for both because as n goes to infinity the n is the denominator for each so I thought they probably each go to zero. but also, the Arg is capitalized here as well, so I am getting the feeling I am doing these wrong. Can anybody help? Thanks!

Answer 1

The meaning of capitalized names such as $\operatorname{Log}$ varies by source. I assume that $ \operatorname{Log}$ has been defined so that it's continuous at $2i$ and at $2$; this is the case for the common definitions I'm familiar with. Check your definition. Then

• $\operatorname{Log}((2/n) + 2i) \to \operatorname{Log}(2i)$ as $n \to \infty$
• $\operatorname{Log}( 2 + 2i/n) \to \operatorname{Log}(2)$ as $n \to \infty$
• $\operatorname{Arg}((1+i)/n) = \operatorname{Arg}(1+i)$ for all $n$; this is a constant sequence. Argument of a complex number is not affected by scaling.
• $(\operatorname{Arg}(1+i))/n \to 0$ since numerator does not depend on $n$, while the denominator grows indefinitely.