Complex Numbers Ques 12
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
$(1980,2 M)$
Show Answer
Answer:
Correct Answer: 12.(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 $
BETA
