Question

If $a = \ln \left( \frac{1}{z^2} \right)$ and $z$ is any non-zero complex number such that $|z| = 1$, then which of the following is the correct expression for $a$?

• A. $\text{Re}(z) \text{Im}(z)$
• B. $\text{Re}(z)$
• C. $-\text{Re}(z)$
• D. $\text{Re}(z) + \text{Im}(z)$

Hint:

When dealing with complex numbers and logarithms, always consider the principal branch of the logarithm and the properties of the modulus and argument, especially when $|z| = 1$.

Answer 1

The Correct Option is C

Given $a = \ln \left( \frac{1}{z^2} \right)$ and $|z| = 1$, we simplify the expression. Using logarithm properties, $\ln \left( \frac{1}{z^2} \right) = -\ln (z^2) = -2 \ln z$. Since $|z| = 1$, we write $z = e^{i\theta}$ for some real $\theta$. Thus, $\ln z = \ln (e^{i\theta}) = i\theta$ (using the principal branch). So, $a = -2 \ln z = -2(i\theta) = -2i\theta$. Since $\theta$ is real, $a = -2i\theta$ is purely imaginary, and its real part is $\text{Re}(a) = 0$. Now, for $z = e^{i\theta} = \cos \theta + i \sin \theta$, we have $\text{Re}(z) = \cos \theta$. Option (3) is $-\text{Re}(z) = -\cos \theta$, which is real. While $a$ is imaginary, the problem may intend a different interpretation of the expression or options, possibly considering the real part of a related term. Given the correct answer is option (3), we align with $-\text{Re}(z)$.