Question
Download Solution PDFमान लीजिए z एक सम्मिश्र संख्या इस प्रकार है जिससे |z| = 4 और \(z = \frac{{5\pi }}{6}\) है। तो z किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
एक सम्मिश्र संख्या के मापांक और तर्क के बीच का संबंध:
किसी सम्मिश्र संख्या \(z\) के लिए यदि मापांक \(|z|\) दिया गया है और तर्क \(\theta\) है, तो निम्नलिखित संबंध सदैव सत्य होता है:
\(z = |z|e^{i\theta}\)
\(e^{i\theta} = \cos \theta+i\sin\theta\)
गणना:
माना कि दी गयी सम्मिश्र संख्या \(z\) है, तो हमारे पास \(|z| = 4\) है और \(z = \dfrac{5\pi}{6}\) है।
इसलिए, \(\theta = \dfrac{5\pi}{6}\).
अब उपरोक्त सूत्र का प्रयोग करने पर,
\(\begin{align*} z &= |z|e^{i\theta}\\ &= 4\left(e^{i\frac{5\pi}{6}}\right)\\ &= 4\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\\ &= 4\left(-\dfrac{\sqrt3}{2} + i\dfrac{1}{2}\right)\\ &= -2\sqrt3+2i \end{align*}\)
अतः आवश्यक सम्मिश्र संख्या \(z = -2\sqrt3+2i\) है।
Last updated on May 30, 2025
->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.
-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.