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I'm rather comfortable with $\varepsilon$-$N$ proofs in real analysis, but I'm taking a complex analysis course and (even in real analysis) the $\varepsilon$-$\delta$ proofs still confuse me a bit.

So we're told to prove that $\lim_{z\to z_0}\mathrm{Re}(z)=\mathrm{Rz}(z_0)$.

The proof goes as follows:

Let $z=x+iy$ and $z_0=x_0+iy_0$. Then $\mathrm{Re}(z) = x$, and $\mathrm{Re}(z_0)=x_0$, and $$\left|\mathrm{Re}(z)-\mathrm{Re}(z_0)\right|= \left|x-x_0\right|<\epsilon$$ If $\delta=\epsilon$, then $\left|\mathrm{Re}(z)-\mathrm{Re}(z_0)\right|<\epsilon$ when $\left|z-z_0\right|<\delta$, $\forall \epsilon > 0$.

To me, this doesn't seem to actually be proving anything. Is there a (hidden) assumption that $x\to z$ and $x_0\to z_0$?

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  • $\begingroup$ Notice how the absolute value of the difference of two complex numbers is greater than or equal to the difference between their real parts (triangle inequality or pythagorean theorem can be used to prove this). This geometric fact is essential to the proof. $\endgroup$ – Cicero Sep 28 '15 at 2:31
  • $\begingroup$ You use $x,x_0$ to denote the real parts, so what would you think $x→ z$ means? (I don't know) $\endgroup$ – Calvin Khor Sep 28 '15 at 2:36
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Note that $$|z - z_0| = |(x-x_0) + i(y-y_0)| = \sqrt{(x-x_0)^2 + (y-y_0)^2} <\delta$$ implies that $$|x-x_0| =\sqrt{(x-x_0)^2} < \delta.$$ What this shows is that if we choose $\delta = \epsilon$ (for whatever $\epsilon>0$ was given to us) then $|z-z_0| < \delta$ implies that $|\text{Re}(z)-\text{Re}(z_0)| = |x-x_0|<\epsilon$.

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You are asked to prove $\lim_{z→z_0}\Re z\ = \Re z_0$. With the notation $z =x+iy = x(z) + iy(z)$, $z_0 = x_0 + iy_0$ (i.e. $x(z) = \Re z$), this is $$\lim_{z→z_0}x(z)\ = x_0$$

We wish to show that:

given any $ε>0$, we can find $δ>0$ such that if $|z-z_0|<δ$, then $|x(z) - x_0| < ε $.

Note the similarity to the $ε-N$ definition:

given any $ε>0$, we can find $N>0$ such that if $n>N$, then $|x(n) - x_0| < ε $.

Conceptually very similar, where "$n$ is large enough" is replaced with "$z$ is close enough to $z_0$".

Anyway. Lets see where the definition takes us, and decide on what $δ$ to use later.

If $|z-z_0| = \sqrt{|x-x_0|^2 + |y-y_0|^2}<\delta$, we know that $$|x-x_0| = \sqrt{|x-x_0|^2} \leq \sqrt{|x-x_0|^2 + |y-y_0|^2} = |z-z_0| <δ $$

We want this last bound to be $ε$. So a good choice of $δ$ is…

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The proof goes as follows:

No it doesn't. Not quite as you've quoted. Just as with $\varepsilon-N$ proofs, you are given an arbitrary $\varepsilon$, you pick a $\delta$ or $N$ based on this $\varepsilon$, and show that if the $\delta / N$ condition holds, then the $\varepsilon$ condition holds. So the proof actually is this:

For any $\varepsilon > 0$, let $\delta = \varepsilon$. If $| z - z_0 | < \delta$, then (as Alex G. has said) $$| x - x_0 | \le \sqrt{(x - x_0)^2 + (y - y_0)^2} = |z - z_0| < \delta = \varepsilon$$ So $\lim\limits_{z \to z_0} \text{Re}(z) = \text{Re}(z_0)$.

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