# limits in complex plane

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• Jan 28th 2013, 03:37 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, 05:40 PM
chiro
Re: limits in complex plane
Hey arlingtonbassett.

What kind of convergence are you talking about (point-wise for example)?
• Jan 28th 2013, 11:52 PM
hollywood
Re: limits in complex plane
I think you could make the argument directly. If $\displaystyle z \ne 0$, then a small enough $\displaystyle \epsilon$-ball around z will have a small range of $\displaystyle \theta$ (where $\displaystyle z=re^{i\theta}$). Specifically, if $\displaystyle |z_n-z|<\epsilon$, then $\displaystyle |\theta_n-\theta| < \frac{\epsilon}{r}$.

- Hollywood