Question

Find the least positive value of n, if $(\frac{1+i}{1-i}{)}^n=1$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

We have,

$\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{(1+i{)}^2}{1-i^2}=\frac{1+2i+i^2}{1-i^2}=\frac{1+2i-1}{1+1}=i$

Therefore,

$(\frac{1+i}{1-i}{)}^n=1$

$i^n=1$

$⟹$ n is a multiple of 4.

The smallest positive value of n is 4.