$$\begin{aligned} z &= \left| z \right| \cdot e^{i \cdot \text{Arg}(z)}, \qquad\qquad\qquad\text{in polar form} \\ \text{Log}(z) &= \ln{\left| z \right|} + i \cdot \text{Arg}(z), \qquad\quad\quad\text{the principal branch of logarithm} \\ \text{log}(z) &= \ln{\left| z \right|} + i\left(\text{Arg}(z) + 2k\pi\right), \quad\,\,\text{where } k \in \mathbb{Z} \\ \text{log}(z) &= \ln{\left| z \right|} + i\cdot\text{arg}(z), \qquad\quad\quad\, \text{a multi-valued function} \\ \end{aligned}$$

Observe Log($z$):

$$\begin{aligned} z &\mapsto \left| z \right|, \qquad\quad\,\,\,\text{continuous in }\mathbb{C} \\ z &\mapsto \ln{\left| z \right|}, \qquad\,\,\;\text{continuous in }\mathbb{C} \setminus \{0\} \\ z &\mapsto \text{Arg}(z), \qquad\text{continuous in }\mathbb{C} \setminus (-\infty,0] \\ \end{aligned}$$

Recall $-\pi \lt \text{Arg}(z) \le \pi$. So as $z$ approaches the x-axis to the left of the origin, it would result in $\pi$ for Arg($z$) if from above, but $-\pi$ if from below.

$$\begin{aligned} \therefore \text{Log}(z) \text{ is continuous in } \mathbb{C} \setminus (-\infty,0]. \end{aligned}$$

Similar line of arguments can be used to reason that the domain of $z^2$ is continuous on $\mathbb{C}$, but the domain of $\sqrt{z}$ is only continuous on $\mathbb{C} \setminus (-\infty,0]$. Do you see how ?

Quiz: what is $\displaystyle\text{Log}(-e^xi)$ ?

Observe $\displaystyle z = -e^xi \,$ lies on the imaginary axis and therefore $\displaystyle\text{Arg}(z) = -\frac{\pi}{2}$

$$\begin{aligned} -e^x i &= e^x(0 - i) = e^x e^{-\frac{\pi}{2}i} \\ \text{Log}(-e^xi) &= x -\frac{\pi}{2}i \\ \end{aligned}$$

$\Box$

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:16 AM