A) \[x=4\,n\], where n is any positive integer
B) \[x=2\,n\], where n is any positive integer
C) \[x=4\,n+1\], where n is any positive integer
D) \[x=2\,n+1\], where n is any positive integer
Correct Answer: A
Solution :
\[{{\left( \frac{1+i}{1-i} \right)}^{x}}={{\left[ \frac{(1+i)(1+i)}{(1-i)(1+i)} \right]}^{x}}\] \[={{\left[ \frac{{{(1+i)}^{2}}}{1-{{i}^{2}}} \right]}^{x}}={{\left[ \frac{1-1+2\,i}{2} \right]}^{x}}\] \[\Rightarrow \] \[{{\left( \frac{1+i}{1-i} \right)}^{x}}={{(i)}^{x}}=1\] (given) \[\Rightarrow \] \[{{(i)}^{x}}={{(i)}^{4n}}\] where, n is any positive integer. \[\Rightarrow \] \[x=4n\] NOTE 4 is the least value for which \[{{i}^{4}}=1\].
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