Question

Argument of \(\frac {1-i√3}{1+i√3}\) is

• A. 60°
• B. 210°
• C. 120°
• D. 240°

Answer 1

The Correct Option is D

Let z = \(\frac {1−i \sqrt{3}}{1+i \sqrt{3}}\)
z = \(\frac {1−i \sqrt{3}}{1+i \sqrt{3}}\) ​ x \(\frac {1−i \sqrt{3}}{1-i \sqrt{3}}\)
z = \(\frac {(1−i \sqrt{3})^2}{1+3}\)
z = \(\frac {-2−2i \sqrt{3}}{4}\)
z = \(-\frac {1}{2} - \frac {i\sqrt {3}}{2}\)
Now r = \(\sqrt {(-\frac {1}{2})^2 + (-\frac {\sqrt{3}}{2})^2 }\)
r = 1
On comparing with z = r cos α + ir sin α
r cos α=​ \(-\frac {1}{2}\)
cos α =​ \(-\frac {1}{2}\)
cos α  = cos 240^o
α = 240^o
Therefore the correct option is (D) 240^o

Answer 2

Steps to solve:
1. Simplify the expression:
  \[  \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}\]
To simplify, multiply the numerator and the denominator by the conjugate of the denominator:
 \[  \frac{(1 - i\sqrt{3})(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})}\]
2. Compute the numerator:
 \[(1 - i\sqrt{3})(1 - i\sqrt{3}) = 1 - 2i\sqrt{3} + (i\sqrt{3})^2 = 1 - 2i\sqrt{3} - 3 = -2 - 2i\sqrt{3}\]
3. Compute the denominator:
\[ (1 + i\sqrt{3})(1 - i\sqrt{3}) = 1^2 - (i\sqrt{3})^2 = 1 - (-3) = 1 + 3 = 4 \]
4. Combine the results:
 \[ \frac{-2 - 2i\sqrt{3}}{4} = \frac{-2}{4} + \frac{-2i\sqrt{3}}{4} = -\frac{1}{2} - \frac{i\sqrt{3}}{2}\]
 This can be written as:
\[-\frac{1}{2}(1 + i\sqrt{3})\]
5. Determine the argument:
 The complex number can be written in polar form as:
\[ z = re^{i\theta} \]
 where \(r\) is the magnitude and \(\theta\) is the argument. For \(-\frac{1}{2}(1 + i\sqrt{3})\), we first find the argument of \(1 + i\sqrt{3}\).
The argument of \(1 + i\sqrt{3}\) is:
\[\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = 60^\circ\]
Since the complex number is multiplied by \(-1\), this rotates the argument by \(180^\circ\):
\[\text{Argument} = 60^\circ + 180^\circ = 240^\circ\]
Final Result:
The argument of the complex number \(\frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}\) is \(240^\circ\).
Therefore, the correct option is:
\[ \boxed{240^\circ} \]