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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.

2 Answers 2

17

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}. $

  • 0
    +1. Modified a denominator to make more explicit the corresponding step (if it does not suit you, please cancel it).2012-11-24
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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 $\,t\,$