1. ## complex limit

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

2. ## Re: complex limit

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

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

Tonio

3. ## Re: complex limit

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

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

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

4. ## Re: complex limit

If i want to show that
$\displaystyle lim_{|z|\rightarrow 0} |\frac{ln(z)}{z^2+a^2}| = 0$
does it get easier?

5. ## Re: complex limit

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

Why it isn't
$\displaystyle \,z^2+a^2\xrightarrow [|z|\to 0]{}{} a^2$ ?
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