Question
Answers

Find the least positive integer value of n, for which \[{{\left( \dfrac{1-i}{1+i} \right)}^{n}}\], where \[i=\sqrt{-1}\], is purely imaginary with the positive imaginary part.

Answer
VerifiedVerified
158.1k+ views
Hint: First of all, simplify the given expression by multiplying (1 – i) to both numerator and denominator and substituting \[{{i}^{2}}=-1\]. Then substitute different positive integral values of n starting from 1 to get desirable value.

Complete step-by-step answer:
We are given an expression \[{{\left( \dfrac{1-i}{1+i} \right)}^{n}}\], we have to find the least positive integer value of n, such that the expression is purely imaginary with the positive imaginary part.
Let us consider the expression given in the question as
\[E={{\left( \dfrac{1-i}{1+i} \right)}^{n}}\]
First of all, let us simplify this expression. By multiplying (1 – i) to both numerator and denominator inside the bracket, we get,
 \[E={{\left[ \dfrac{\left( 1-i \right)\left( 1-i \right)}{\left( 1+i \right)\left( 1-i \right)} \right]}^{n}}\]
We know that \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]. By applying this in the denominator of the above expression, we get,
\[E={{\left[ \dfrac{{{\left( 1-i \right)}^{2}}}{\left( {{1}^{2}}-{{\left( i \right)}^{2}} \right)} \right]}^{n}}\]
We also know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. By applying this in the numerator of the above expression, we get,
\[E={{\left( \dfrac{1+{{\left( i \right)}^{2}}-2i}{1-{{\left( i \right)}^{2}}} \right)}^{n}}\]
We know that \[i=\sqrt{-1}\]
By squaring both sides, we get,
\[{{i}^{2}}=-1\]
By substituting \[{{i}^{2}}=-1\] in the above expression, we get,
\[E={{\left( \dfrac{1-1-2i}{1-\left( -1 \right)} \right)}^{n}}\]
We get, \[E={{\left( \dfrac{-2i}{2} \right)}^{n}}\]
Or, \[E={{\left( -i \right)}^{n}}\]
Now, we have to find the least positive integer value of n such that the above expression is purely imaginary with a positive imaginary part.
Let us substitute n = 1 in the above expression, we get
\[E={{\left( -i \right)}^{1}}=-i\]
‘\[E=-i\]’ is purely imaginary but its imaginary part is not possible, so \[n\ne 1\] .
Now, let us substitute n = 2, we get
\[E={{\left( -i \right)}^{2}}={{\left( i \right)}^{2}}\]
\[E=-1\]
\[E=-1\] is not imaginary, so \[n\ne 2\].
Now, let us substitute n = 3, we get,
\[E={{\left( -i \right)}^{3}}=-{{\left( i \right)}^{3}}\]
We know that, \[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]. By applying this, we can write the above expression as,
\[E=-\left( {{\left( i \right)}^{2}}.{{\left( i \right)}^{1}} \right)\]
We know that, \[{{i}^{2}}=-1\]. Therefore we get,
\[E=-\left( -1 \right)i=i\]
E = i is purely imaginary and its imaginary part is positive as well, so we get n = 3.
Hence, we get the least positive integer value of n = 3 for which \[{{\left( \dfrac{1-i}{1+i} \right)}^{n}}\] is purely imaginary with a positive imaginary part.

Note: Some students start substituting n = 1, 2, 3 in the original expression that is in \[{{\left( \dfrac{1-i}{1+i} \right)}^{n}}\] only. But it is advisable to first simplify the expression and then only substitute the values of n as the question becomes very lengthy in the previous case. Also, after giving \[E={{\left( -i \right)}^{n}}\], students often make mistakes and get the answer n = 1 but they must take care that they not only need to get purely imaginary expression but it should be positive as well.