# complex limit

• June 16th 2011, 02:30 PM
hurz
complex limit
Show that
$lim_{|z|\rightarrow 0} \frac{ln(z)}{z^2+a^2} = 0$
where z is a complex number
Thanks
• June 16th 2011, 03:48 PM
tonio
Re: complex limit
Quote:

Originally Posted by hurz
Show that
$lim_{|z|\rightarrow 0} \frac{ln(z)}{z^2+a^2} = 0$
where z is a complex number
Thanks

Use L´Hospital's rule , as $Ln z=\ln |z|+i\arg (z) \xrightarrow [|z|\to 0] . -\infty \, , \,z^2+a^2\xrightarrow [|z|\to 0]{}{} \infty$

Tonio
• June 16th 2011, 04:02 PM
hurz
Re: complex limit
Quote:

Originally Posted by tonio
Use L´Hospital's rule , as $Ln z=\ln |z|+i\arg (z) \xrightarrow [|z|\to 0] . -\infty \, , \,z^2+a^2\xrightarrow [|z|\to 0]{}{} \infty$

Tonio

I'm sorry, i'm not getting why
$\,z^2+a^2\xrightarrow [|z|\to 0]{}{} \infty$.

Why it isn't
$\,z^2+a^2\xrightarrow [|z|\to 0]{}{} a^2$ ?
In the cause above, i could not use L'Hospital's Rule.
Regards
• June 16th 2011, 04:12 PM
hurz
Re: complex limit
If i want to show that
$lim_{|z|\rightarrow 0} |\frac{ln(z)}{z^2+a^2}| = 0$
does it get easier?
• June 16th 2011, 06:41 PM
tonio
Re: complex limit
Quote:

Originally Posted by hurz
I'm sorry, i'm not getting why
$\,z^2+a^2\xrightarrow [|z|\to 0]{}{} \infty$.

Why it isn't
$\,z^2+a^2\xrightarrow [|z|\to 0]{}{} a^2$ ?
In the cause above, i could not use L'Hospital's Rule.
Regards

Of course you´re right, my bad. Mistook the limit to infinity instead of zero in the denominator.

Tonio