Thread: 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

Hey arlingtonbassett.

What kind of convergence are you talking about (point-wise for example)?

I think you could make the argument directly. If , then a small enough -ball around z will have a small range of (where ). Specifically, if , then .

- Hollywood