Question
Download Solution PDFThe smallest positive integer n for which \(\left(\frac{1+i}{1 - i}\right)^n = 1\), is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
- The complex unit i is defined as \(\rm i=\sqrt{-1}\).
- i2 = -1, i3 = - i, i4 = 1 etc.
- (a + b)2 = a2 + 2ab + b2
Calculation:
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)
\( \left(\dfrac{1+i}{1-i}\right)^n=\ i^n\ =\ 1\)
The smallest positive integer for which in = 1 is n = 4.
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