Question
Download Solution PDFसबसे छोटा धनात्मक पूर्णांक n क्या है जिसके लिए \(\left(\dfrac{1+i}{1-i}\right)^n=1\) है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
- संयुक्त इकाई i को \(\rm i=\sqrt{-1}\) के रूप में परिभाषित किया गया है।
- i2 = -1, i3 = - i, i4 = 1 इत्यादि
- (a + b)2 = a2 + 2ab + b2
गणना:
Given that,
\(\left(\frac{1+i}{1 - i}\right)^n = 1\) ---(1)
Let, y = \(\left(\frac{1+i}{1 - i}\right)\)
\(\Rightarrow \ y=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}\).
\(\Rightarrow\ y\ =\dfrac{(1+i)^2}{1^2-i^2}\)
∵ (a + b)2 = a2 + 2ab + b2
\(\Rightarrow\ y\ =\dfrac{1+2i+i^2}{1-i^2}\)
\(\Rightarrow y\ =\dfrac{2i}{2}=i\) (∵ i2 = -1)
Therefore, from equation (1)
\( \left(\dfrac{1+i}{1-i}\right)^n=\ i^n\ =\ 1\)
सबसे छोटा धनात्मक पूर्णांक n = 4 है, जिसके लिए in = 1 है।
Last updated on May 31, 2025
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