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Im trying to find the limit of:

$$ \frac{\operatorname{Re}(z) \operatorname{Im}(z)}{z^2}$$

as z tends to zero.

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2 Answers 2

up vote 14 down vote accepted

It doesn't exist since it can give different results depending on the direction. For example for real $t$

$$ \lim_{t\to 0} \frac{\operatorname{Re}(t) \operatorname{Im}(t)}{t^2} = \lim_{t \to 0}\frac{t \cdot 0}{t^2} = 0 $$

but

$$ \lim_{t\to 0} \frac{\operatorname{Re}(t + i t) \operatorname{Im}(t + i t)}{(t+it)^2} = \lim_{t \to 0} \frac{t \cdot t}{2 i \, t^2} = \frac{-i}{2}. $$

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+1. Modified a denominator to make more explicit the corresponding step (if it does not suit you, please cancel it). –  Did Nov 24 '12 at 12:10

Another approach: use polar coordinates

$$z=re^{it}\,\,,\,0\leq t\leq 2\pi\Longrightarrow Re(z)=r\cos t\,\,,\,Im(z)=r\sin t\,\,,\,\text{and}\,\,\,z\to 0\Longleftrightarrow r\to 0 \Longrightarrow$$

$$\Longrightarrow \frac{Re(z)Im(z)}{z^2}=\frac{r^2\cos t\sin t}{r^2(\cos^2 t-\sin^2 t+2i\cos t\sin t)}=\frac{1}{2}\frac{\sin 2t}{\cos 2t+i\sin 2t}\xrightarrow [r\to 0]{} \text{doesn't exist}$$

as it depends on the angle $\,t\,$

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