复对数主值方程\(\text{Ln}\,ar{z}= \overline{\text{Ln}(z)}\)求解方法咨询

Hey there, let's work through this problem step by step—you were on the right track but made a small misstep in your reasoning, so let's fix that together!

First, let's recall the definition of the principal branch of the complex logarithm (denoted (\text{Ln})):
For any non-zero complex number (z = |z|e^{i\arg z}) (where the principal argument (\arg z) lies in the interval ((-\pi, \pi])), the principal logarithm is:
$$\text{Ln}(z) = \ln|z| + i\arg z$$
Here, (\ln|z|) is the real-valued natural logarithm of the modulus of (z), and (i\arg z) accounts for the imaginary part from the argument.

Now let's break down both sides of your equation:

Left-hand side: (\text{Ln}(\bar{z})) #

The conjugate of (z), (\bar{z}), has the same modulus as (z) (so (|\bar{z}| = |z|)), and its principal argument is:

• If (\arg z \in (-\pi, \pi)), then (\arg(\bar{z}) = -\arg z) (conjugating flips the sign of the argument)
• If (\arg z = \pi) (i.e., (z) is a negative real number), then (\bar{z} = z), so (\arg(\bar{z}) = \pi)

So substituting into the logarithm definition:
$$\text{Ln}(\bar{z}) = \ln|\bar{z}| + i\arg(\bar{z}) = \ln|z| + i\arg(\bar{z})$$

Right-hand side: (\overline{\text{Ln}(z)}) #

Taking the conjugate of (\text{Ln}(z)) (which is a complex number) flips the sign of the imaginary part:
$$\overline{\text{Ln}(z)} = \overline{\ln|z| + i\arg z} = \ln|z| - i\arg z$$
(Note: (\ln|z|) is real, so its conjugate is itself.)

Set the two sides equal and simplify #

Now equate the two expressions:
$$\ln|z| + i\arg(\bar{z}) = \ln|z| - i\arg z$$
Subtract (\ln|z|) from both sides, then divide by (i):
$$\arg(\bar{z}) = -\arg z$$

Analyze when this holds #

Let's split into cases:

• Case 1: (z) is not a negative real number ((\arg z \in (-\pi, \pi))):
  Here, (\arg(\bar{z}) = -\arg z) is always true. The modulus (|z|) can be any positive real number—your earlier step setting (\ln|z|=1) was unnecessary, since the modulus terms cancel out entirely with no restrictions.
• Case 2: (z) is a negative real number ((\arg z = \pi)):
  Here, (\arg(\bar{z}) = \pi), but (-\arg z = -\pi). Since (\pi \neq -\pi), the equation does not hold for negative real numbers.

Final conclusion #

The equation (\text{Ln},\bar{z}= \overline{\text{Ln}(z)}) is satisfied by all non-zero complex numbers except negative real numbers. In other words, the solution set is (\mathbb{C} \setminus { x \in \mathbb{R} \mid x < 0 }).

备注：内容来源于stack exchange，提问作者Jenn