If a complex-valued function converges, then its real part and imaginary part converge.

232 Views Asked by At

Let $\mathcal R$ and $\mathcal I$ denote the real and imaginary part of a complex-valued function $f: \mathbb{R} \to \mathbb C$.

Is the following claim true?

Let $f_n = (\mathcal{R}_n + i \mathcal{I}_n)$, then $f_n$ converges to $f = \mathcal{R} + i \mathcal{I}$ if and only if $\mathcal{R}_n \to \mathcal{R}$ and $\mathcal{I}_n \to \mathcal{I}$.

I think it is true because it follows the fact that (z_n) Converges IFF (Re(z_n)) and (Im(z_n)) Converges

At each point of the complex-valued function value converges iff the real part and the imaginary part at that particular point converge.

Then, we can use it to conclude that if a complex-valued function is pointwise convergent iff the real part and imaginary part is pointwise convergent.

Is my reasoning correct?

Any thoughts are appreciated!

1

There are 1 best solutions below

0
On

Yes, this is true. The imaginary part function is continuous, so if $f_n$ converges than so does $im(f_n)$ (and same for the real part).

The other direction requires a bit of $\delta-\epsilon$ but is pretty straightforward to prove.