Question
Download Solution PDFसमीकरण \({\left( {\frac{{\sqrt 2 + i\;\sqrt 2 }}{2}} \right)^{64}}\) का मूल्यांकन कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)
गणना:
दिए गए समीकरण:\({\left( {\frac{{√ 2 + i\;√ 2 }}{2}} \right)^{64}}\) को निम्न रूप में पुनः लिखा जा सकता है
= \((\frac{\sqrt 2}{2}\ +\ \frac{\sqrt 2}{2}i)^{64}\)
= \((\frac{1}{\sqrt 2}\ +\ \frac{1}{\sqrt 2}i)^{64}\)
चूँकि हम जानते हैं कि sin π/4 = 1/√2 = cos π/4
इसलिए, हम दिए गए समीकरण\({\left( {\frac{{√ 2 + i\;√ 2 }}{2}} \right)^{64}}\)को निम्न रूप में लिख सकते हैं
= (cos π/4 + i sin π/4)64
चूँकि हम जानते हैं कि, \({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)
⇒ (cos π/4 + i sin π/4)64 = (ei ⋅ π/4 )64
⇒ (cos π/4 + i sin π/4)64 = ei ⋅ 16π
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