# limits in complex plane

• Jan 28th 2013, 04:37 PM
arlingtonbassett
limits in complex plane
If z_n = r_n e^{ix_n} converges to z = re^{ix}, one may ask why is it that x_n converges to x? One can show this directly using the fact that the (z_n) sequence converges iff its real and imaginary components converge. It is fairly obvious that r_n converges to r (use ||x|-|y||<=|x-y|). Using e^{ix} = cosx + isinx one can get an inequality involving sin(x_n), cos(x_n), cosx, and sinx, then an inequality in x_n,x alone that shows x_n converges to x.
arlington
• Jan 28th 2013, 06:40 PM
chiro
Re: limits in complex plane
Hey arlingtonbassett.

What kind of convergence are you talking about (point-wise for example)?
• Jan 29th 2013, 12:52 AM
hollywood
Re: limits in complex plane
I think you could make the argument directly. If $z \ne 0$, then a small enough $\epsilon$-ball around z will have a small range of $\theta$ (where $z=re^{i\theta}$). Specifically, if $|z_n-z|<\epsilon$, then $|\theta_n-\theta| < \frac{\epsilon}{r}$.

- Hollywood