Question

If ${\left[\frac{1+i}{1-i}\right]}^m=1$, then find the least positive integral value of m.

Answer 1

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

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

$\therefore {\left[\frac{1+i}{1-i}\right]}^m=(i{)}^m=1$

$\therefore (i{)}^m=1=(i{)}^4$

$\therefore m=4$