• # question_answer If ${{\left( \frac{1+i}{1-i} \right)}^{m}}=1,$then the least integral value of $m$ is [IIT 1982; MNR 1984; UPSEAT 2001; MP PET 2002] A) 2 B) 4 C) 8 D) None of these

$\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{{{(1+i)}^{2}}}{2}=\frac{2i}{2}=i$ $\therefore$${{\left( \frac{1+i}{1-i} \right)}^{m}}={{i}^{m}}=1$ (as given) So the least value of $m=4$$\{\because {{i}^{4}}=1\}$