Question

The imaginary part of the complex number \( \log(1+i) \) is

• A. \(\pi/4\)
• B. \(\pi/2\)
• C. \(3\pi/2\)
• D. \(3\pi/4\)

Hint:

To find the logarithm of a complex number \(z=x+iy\), first convert it to polar form \(z=re^{i\theta}\), where \(r=\sqrt{x^2+y^2}\) and \(\theta=\arctan(y/x)\). Then, \( \text{Log}(z) = \ln(r) + i\theta \). The imaginary part is simply the principal argument \(\theta\).

Answer 1

The Correct Option is A

Step 1: Convert the complex number \(z = 1+i\) into polar form \(z = re^{i\theta}\).
The modulus is \( r = |z| = \sqrt{1^2+1^2} = \sqrt{2} \). The argument is \( \theta = \arctan(\frac{1}{1}) = \frac{\pi}{4} \) (since the point (1,1) is in the first quadrant). So, \( 1+i = \sqrt{2} e^{i\pi/4} \).

Step 2: Apply the complex logarithm formula.
The principal value of the complex logarithm is given by \( \text{Log}(z) = \ln(r) + i\theta \), where \( -\pi < \theta \le \pi \). \[ \text{Log}(1+i) = \ln(\sqrt{2}) + i\frac{\pi}{4} \]

Step 3: Identify the real and imaginary parts.
From the expression above, the real part is \( \ln(\sqrt{2}) = \frac{1}{2}\ln(2) \) and the imaginary part is \( \frac{\pi}{4} \).