Complex analysis (basic, no solutions)

Question

I have to use only the following definition to prove the limit exists.

Using def (2) sec 15. prove the following.

Section 15 (2) =$ |f(z) - w_0 |< \epsilon \space \space \space $ Whenever $ 0< | z-z_0 | < \delta $

Let a,b,c denote complex constants then using def (2) sec 15. prove the following.

1)$\lim_{z \to z_{0}} (az+b) = (az_0 +b)$

$\lim_{z \to z_{0}} |(az+b)- (az_0 +b)| = |az- az_0|= |a| |z-z_0 |< \epsilon $

Let $\delta = \frac {\epsilon}{|a| }$ I feel like i need to say something here to complete this proof?

2) $\lim_{z \to z_{0}} (z^2 +c ) = (z_{0}^{2} +c) $

$\lim_{z \to z_{0}} |(z^2 +c ) -(z_{0}^{2} +c)| = |z^2 -z_{0}^{2}|=|(z -z_{0})(z +z_{0})| = |z -z_{0}||z +z_{0}| < \epsilon $

Either $|z -z_{0}|>|z +z_{0}| $ or $ |z -z_{0}| <|z +z_{0}| $ let $ \delta = \max \{ 2|z -z_{0}|,2|z +z_{0}| \} $

Clearly $ |z -z_{0}| < \delta $

Assuming these are right how do i finish them off? ( I have never taken a class on analysis and im supposed to learn this without ever doing any examples in class for my midterm.)

Answer 1

First one looks good. In the second one your choice of $\delta $ is dependent on the variable $z$, which is incorrect. Your $\delta $ must solely depend on the choice of $\epsilon$. Try to carefully understand the definition of limit of a function.

For the second one you can proceed as follows.

$$ |(z^2 +c ) -(z_{0}^{2} +c)| = |z^2 -z_{0}^{2}|=|(z +z_{0})(z -z_{0})| = |z +z_{0}||z -z_{0}| \le(|z|+|z_0|)|z-z_0|$$

We now make use of the following inequality which is a consequence of the Triangle Inequality, $$||a|-|b||\le|a-b|$$ For $|z-z_0|<1$, we have $$||z|-|z_0||\le|z-z_0|<1\implies|z|-|z_0|<1\implies|z|<1+|z_0|$$

and hence, $$|(z^2 +c ) -(z_{0}^{2} +c)| \le(|z|+|z_0|)|z-z_0|<(1+2|z_0|)|z-z_0|-------(1)$$

Now choose, $\delta=min\left(1,\frac{\displaystyle\epsilon}{\displaystyle1+2|z_0|}\right)$, which does not depend on the variable $z$.

Since, $0<|z-z_0|<\delta\le 1 $. Equation (1) holds, $$|(z^2 +c ) -(z_{0}^{2} +c)| <(1+2|z_0|)|z-z_0|$$

Again, as $\delta \le \frac{\displaystyle\epsilon}{\displaystyle(1+2|z_0|)}$

So for $0<|z-z_0|<\delta$, $$|(z^2 +c ) -(z_{0}^{2} +c)| < (1+2|z_0|)|z-z_0|<(1+2|z_0|)\frac{\displaystyle\epsilon}{\displaystyle(1+2|z_0|)}=\epsilon$$

Q.E.D