Question

Let the two values of \( z = \frac{1 - i}{\sqrt{1 + i}} \) be \( z_1 \) and \( z_2 \). If \( -\frac{\pi}{2}<\text{Arg}(z_1)<\text{Arg}(z_2)<\pi \), then \( \text{Arg}(z_1) + \text{Arg}(z_2) = \)

• A. \( \frac{\pi}{4} \)
• B. \( \frac{3\pi}{2} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{2} \)

Hint:

When dealing with complex numbers raised to fractional powers, remember that there are multiple roots. Express the complex numbers in polar form to easily calculate powers and roots using De Moivre's theorem. Pay attention to the principal argument and the range specified for the arguments of the roots.

Answer 1

The Correct Option is D

First, let's simplify the expression for \( z \): $$ z = \frac{1 - i}{\sqrt{1 + i}} $$ We can write \( 1 + i \) in polar form: \( 1 + i = \sqrt{1^2 + 1^2} e^{i \arctan(1/1)} = \sqrt{2} e^{i \pi/4} \).
Then, \( \sqrt{1 + i} = (\sqrt{2} e^{i \pi/4})^{1/2} = 2^{1/4} e^{i \pi/8} \).
Now, let's write \( 1 - i \) in polar form: \( 1 - i = \sqrt{1^2 + (-1)^2} e^{i \arctan(-1/1)} = \sqrt{2} e^{-i \pi/4} \).
So, \( z = \frac{\sqrt{2} e^{-i \pi/4}}{2^{1/4} e^{i \pi/8}} = 2^{1/2 - 1/4} e^{-i \pi/4 - i \pi/8} = 2^{1/4} e^{-i (2\pi + \pi)/8} = 2^{1/4} e^{-i 3\pi/8} \).
The argument of \( z \) is \( -\frac{3\pi}{8} \).
However, the square root has two values.
Let \( w = 1 + i = \sqrt{2} e^{i (\pi/4 + 2k\pi)} \), where \( k \) is an integer.
Then \( \sqrt{w} = 2^{1/4} e^{i (\pi/8 + k\pi)} \).
For \( k = 0 \), \( \sqrt{w_1} = 2^{1/4} e^{i \pi/8} \).
For \( k = 1 \), \( \sqrt{w_2} = 2^{1/4} e^{i (\pi/8 + \pi)} = 2^{1/4} e^{i 9\pi/8} \).
So the two values of \( z \) are: $$ z_1 = \frac{\sqrt{2} e^{-i \pi/4}}{2^{1/4} e^{i \pi/8}} = 2^{1/4} e^{-i 3\pi/8} $$ $$ z_2 = \frac{\sqrt{2} e^{-i \pi/4}}{2^{1/4} e^{i 9\pi/8}} = 2^{1/4} e^{-i (\pi/4 - 9\pi/8)} = 2^{1/4} e^{-i (2\pi - 9\pi)/8} = 2^{1/4} e^{i 7\pi/8} $$ The arguments are \( \text{Arg}(z_1) = -\frac{3\pi}{8} \) and \( \text{Arg}(z_2) = \frac{7\pi}{8} \).
We have \( -\frac{\pi}{2}<-\frac{3\pi}{8}<\frac{7\pi}{8}<\pi \), which satisfies the given condition.
Then, \( \text{Arg}(z_1) + \text{Arg}(z_2) = -\frac{3\pi}{8} + \frac{7\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2} \).