Show that $x=Re(z)$, and $y=Im(z)$ are not complex differentiable

Question

I am self-learning Gamelin’s Complex Analysis and I have performed a calculation w.r.t ex. 3 on p. 46 and I ended ip misinterpreting my conclusion and now I’d love some feedback.

The exercise is to show from the definition of the complex derivative that the functions $x=Re(z)$, and $y=Im(z)$ are not complex differentiable at any point in $\mathbb{C}$.

I started out like this: $$1+ i0=1=\lim_{z\to z_0} 1=\lim_{z \to z_0} \frac{z-z_0}{z-z_0}=\lim_{z\to z_0} \frac{Re(z)-Re(z_0)}{z-z_0} + i \lim_{z \to z_0}\frac{Im(z) - Im(z_0)}{z-z_0}$$

Identifying real and imaginary parts we obtain $\lim_{z \to z_0} \frac{Re(z)-Re(z_0)}{z-z_0} = 1$, and $\lim_{z \to z_0} \frac{Im(z)-Im(z_0)}{z-z_0} = 0$.

Now, I am unsure what my result means or if it means anything at all?

Thank you in advance, Isak

Answer 1

This answer is a synthesis of the other answers and comments and this synthesis is what I personally think makes up a complete answer to what the OP was asking for.

The functions $x=\mathbb{Re}(z)$ and $y=\mathbb{Im}(z)$ are not complex differentiable at any point in $\mathbb{C}$ because their limits (of the corresponding complex derivative) do not match up at any point in $\mathbb{C}$.

Actually, since $\Delta z \to 0$ along many paths in $\mathbb{C}$ we calculate $\lim_{\Delta z \to 0} \frac{\mathbb{Re}(z+\Delta z) - \mathbb{Re}(z)}{\Delta z}$ and $\lim_{\Delta z \to 0} \frac{\mathbb{Im}(z+\Delta z) - \mathbb{Im}(z)}{\Delta z}$ as $\Delta z = \Delta x$ and when $\Delta z = i\Delta y$.

This gives $$\lim_{\Delta z \to 0} \frac{\mathbb{Re}(z+\Delta z) - \mathbb{Re}(z)}{\Delta z} = \begin{cases} 1,& \text{if } \Delta z = \Delta x\\ 0, & \text{if} \Delta z = i\Delta y \end{cases}$$

and, $$\lim_{\Delta z \to 0} \frac{\mathbb{Im}(z+\Delta z) - \mathbb{Im}(z)}{\Delta z} = \begin{cases} 0,& \text{if } \Delta z = \Delta x\\ -i, & \text{if} \Delta z = i\Delta y \end{cases}$$

So the OP imposes a restriction to the real line when OP is assuming that $\lim(a_n+b_n)= \lim(a_n)+\lim(b_n)$ without knowing that the individual limits exist.

Answer 2

Let $z, h \in \mathbb{C}, h \neq 0$. Then $$\frac{\operatorname{Re}(z+h)- \operatorname{Re}(z)}{h} = \frac{\operatorname{Re}(z) + \operatorname{Re}(h)- \operatorname{Re}(z)}{h} = \frac{\operatorname{Re}(h)}{h}.$$ Now you can look at the sequences $a_n := \frac{1}{n}, b_n := \frac{i}{n} \in \mathbb{C}$ and observe that $$\lim_{n \to \infty} a_n = 0 = \lim_{n \to \infty} b_n$$ but $$\lim_{n \to \infty} \frac{\operatorname{Re}(a_n)}{a_n} = \lim_{n \to \infty} \frac{a_n}{a_n} = 1 \neq 0 = \lim_{n \to \infty} \frac{0}{b_n} = \lim_{n \to \infty} \frac{\operatorname{Re}(b_n)}{b_n},$$ so the limit $\frac{\operatorname{Re}(z+h)- \operatorname{Re}(z)}{h}$ as $h \to 0$ does not exist.

Answer 3

Let $h$ be a real number.

$$\lim_{h\to0}\frac{\Re(z+h)-\Re(z)}h=\lim_{h\to0}\frac{h}h\ne\lim_{h\to0}\frac{\Re(z+ih)-\Re(z)}h=\lim_{h\to0}\frac{0}h$$

and simlarly for $\Im(z)$.