Question:
What is the principal argument of \((-1, -i)\) (i.e. \(-1 – i\)), where \(i = \sqrt{-1}\)?
\((-1 – i)\) का मुख्य कोणांक (principal argument) क्या है, जहाँ \(i = \sqrt{-1}\)?
Options:
- \(\pi/4\)
- \(-\pi/4\)
- \(-3\pi/4\)
- \(3\pi/4\)
- \(\pi/4\)
- \(-\pi/4\)
- \(-3\pi/4\)
- \(3\pi/4\)
Correct Answer: c)
Explanation:
Let \(z = x + iy = -1 – i\).
Here \(x = -1 < 0\) and \(y = -1 < 0\), so the complex number lies in the third quadrant.
The acute angle \(\alpha = \tan^{-1}|y/x| = \tan^{-1}(1) = \frac{\pi}{4}\).
In the third quadrant, the principal argument \(\theta \in (-\pi, \pi]\) is given by:
\[\theta = -(\pi – \alpha) = -\left(\pi – \frac{\pi}{4}\right) = -\frac{3\pi}{4}\]
Hence, the correct option is (c).
माना \(z = -1 – i\)।
यहाँ \(x = -1 < 0\) तथा \(y = -1 < 0\) है, अतः यह तीसरे चतुर्थांश में स्थित है।
न्यूनकोण \(\alpha = \tan^{-1}|-1/-1| = \frac{\pi}{4}\)।
तृतीय चतुर्थांश में मुख्य कोणांक का मान:
\[\theta = -(\pi – \alpha) = -\left(\pi – \frac{\pi}{4}\right) = -\frac{3\pi}{4}\]
अतः सही विकल्प (c) है।