Question

Principal Values of Logarithmic Functions

Case	Complex Number	Principal Value
A	\(\log(-2)\)	\(\log 2 + i\pi\)
B	\(\log(5i)\)	\(\log 5 + \dfrac{i\pi}{2}\)
C	\(\log(\sqrt{3}-i)\)	\(\log 2 - \dfrac{i\pi}{6}\)
D	\(\log(2-3i)\)	\(\dfrac{1}{2}\log 13 - i\tan^{-1}\left(\dfrac{3}{2}\right)\)

 

• A. A, B
• B. B, C
• C. B, D
• D. A, D

Hint:

For complex logarithm, first find modulus and principal argument: \(\log z=\log r+i\theta\).

Answer 1

The Correct Option is C

Concept:
For a complex number: \[ z=re^{i\theta} \] the principal value of logarithm is: \[ \log z=\log r+i\theta \] where \(\theta\) is the principal argument.

Step 1: Check statement B. \[ 5i=5e^{i\pi/2} \] So: \[ \log(5i)=\log5+\frac{i\pi}{2} \] Thus, B is correct.

Step 2: Check statement C. \[ \sqrt{3}-i \] Its modulus is: \[ r=\sqrt{3+1}=2 \] Its argument is: \[ -\frac{\pi}{6} \] So: \[ \log(\sqrt{3}-i)=\log2-\frac{i\pi}{6} \] Therefore, C is incorrect.

Step 3: Check statement D. \[ 2-3i \] Modulus: \[ r=\sqrt{2^2+(-3)^2}=\sqrt{13} \] Argument: \[ \theta=-\tan^{-1}\frac{3}{2} \] Therefore: \[ \log(2-3i)=\frac{1}{2}\log13-i\tan^{-1}\frac{3}{2} \] So, D is correct.

Step 4: Final answer.
The correct statements are: \[ B,D \] \[ \therefore \text{Correct Answer is (C)} \]