Complex Numbers 1 Question 9
9.
The smallest positive integer $n$ for which $(\frac{1+i}{1-i})^n=1$, is
(a) 8
(b) 16
(c) 12
(d) None of these
$(1980,2 M)$
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Answer:
Correct Answer: 9. (d)
Solution:
- Since, $(\frac{1+i}{1-i})^{n}=1 \Rightarrow (\frac{1+i}{1-i} \times \frac{1+i}{1+i})^{n}=1$
$ \begin{aligned} \Rightarrow & & \frac{2 i^{n}}{2} & =1 \\ \Rightarrow & & i^{n} & =1 \end{aligned} $
The smallest positive integer $n$ for which $i^{n}=1$ is 4 .
$ \therefore \quad n=4 $
