2
$\begingroup$

How do I prove that $\lim_{z\to0}\frac{\Re(z)\Im(z)}{|z|}=0? $ I've tried it using $\varepsilon-\delta$ language, but can't really get anywhere.

Any hints would be greatly appreciated.

Thanks.

  • 0
    The same limit came up in another question: http://math.stackexchange.com/questions/26672/finding-limit-of-function-with-more-than-one-variable2011-04-16

2 Answers 2

10

Clearly neither Re(z) nor Im(z) can have magnitude larger than |z|. So:

$\displaystyle{ \lim_{z\to0}\left|\frac{\mathrm{Re}(z)\mathrm{Im}(z)}{|z|}\right| = \lim_{z\to0}\frac{|\mathrm{Re}(z)|\times|\mathrm{Im}(z)|}{|z|} \le \lim_{z\to0}\frac{|z|\times|z|}{|z|} = \lim_{z\to0}|z| = 0 }$

Since the magnitude goes to zero, the quantity goes to zero.

7

If we rewrite it

$\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2+y^2}}$

does it look more familiar? You can use polar coordinates or Young's inequality.

  • 2
    @Gerben: To say the truth it's just my laziness that made me mention Young's inequality. The inequality you need here is $|xy|\le \frac{1}{2}(|x|^2+|y|^2)$ which you can prove in a nanosecond by observing that $(\lvert x \rvert-\lvert y \rvert)^2\ge 0$. Sure, it is a special case of Young, but it's much easier.2011-04-16