Compute the given complex limit of limz→i(ln|x2+y2|+iarctan(yx))\lim_{z \rightarrow i}(\ln|x^2+y^2| + i\arctan(\frac{y}{x}))

I can see that this reduces to lim(x,y)→(0,1)(ln|x2+y2|+iarctan(yx))\lim_{(x,y) \rightarrow (0,1)} (\ln|x^2+y^2|+ i\arctan(\frac{y}{x})). The real part of the complex expression is 00, but I’m having trouble understanding what I would do for the complex part. This seems to be just θ\theta as the angle of ii which would be π/2\pi/2, but am I missing something here?

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2 Answers

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No indeterminated form because ln|x2+y2|+iarctan(yx)=ln|z|2+iargz\ln|x^2+y^2| + i\arctan(\frac{y}{x})=\ln|z|^2+i \arg z so when z→iz\to i you have directly limz→i(ln|z|2+iargz)=(ln1+iπ2)=iπ2\lim_{z\to i}(\ln|z|^2+i \arg z)=(\ln 1+i \frac{\pi}{2})=\frac{i\pi}{2}

As you found out yourself the problem is with

lim(x,y)→(0,1)arctanyx .\lim_{(x,y)\to(0,1)}\arctan{y\over x}\ .

This limit does not exist, since

limx→0+arctan1x=π2,limx→0−arctan1x=−π2 .\lim_{x\to0+}\arctan{1\over x}={\pi\over2},\qquad \lim_{x\to0-}\arctan{1\over x}=-{\pi\over2}\ .