Question

If ${{\left( \dfrac{1+i}{1-i} \right)}^{\dfrac{m}{2}}}={{\left( \dfrac{1+i}{1-i} \right)}^{\dfrac{n}{3}}}=1,\,(m,n\in N)$ then the greatest common divisor of the least values of $m$and $n$ is.

Answer 1

Hint: In order to find the greatest common divisor, we will rationalize each of the both equations and determine the values of $m$ and $n$ by multiplying and dividing the equation by its conjugate.
The greatest common divisor of two or more integers is the greatest number which divides each of the integers.

Formula Used: The value of \[{{i}^{4}}=1\] and \[{{i}^{2}}=-1\].
The expansion is:
$\begin{align}
  & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\
 & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
\end{align}$

Complete step by step solution: We are given ${{\left( \dfrac{1+i}{1-i} \right)}^{\dfrac{m}{2}}}={{\left( \dfrac{1+i}{1-i} \right)}^{\dfrac{n}{3}}}=1,\,(m,n\in N)$ and we have to find the greatest common divisor of the least values of $m$and $n$.
First we will rationalize equation ${{\left( \dfrac{1+i}{1-i} \right)}^{\dfrac{m}{2}}}=1$ by multiplying and dividing it to its conjugate. The conjugate of this equation will be $\left( \dfrac{1+i}{1+i} \right)$.
$\begin{align}
  & {{\left( \dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i} \right)}^{\dfrac{m}{2}}}=1 \\
 & {{\left( \dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1+i \right)\left( 1-i \right)} \right)}^{\dfrac{m}{2}}}=1
\end{align}$
Now we will expand using the formula,
${{\left( \dfrac{1+{{i}^{2}}+2i}{1-{{i}^{2}}} \right)}^{\dfrac{m}{2}}}=1$
Substituting the value \[{{i}^{2}}=-1\].
${{\left( \dfrac{1-1+2i}{1-\left( -1 \right)} \right)}^{\dfrac{m}{2}}}=1$
We will now simplify the expression,

$\begin{align}
  & {{\left( \dfrac{2i}{2} \right)}^{\dfrac{m}{2}}}=1 \\
 & {{\left( i \right)}^{\dfrac{m}{2}}}=1 \\
 & {{\left( i \right)}^{\dfrac{m}{2}}}={{i}^{4}} \\
 & \dfrac{m}{2}=4 \\
 & m=8
\end{align}$
Now we will rationalize ${{\left( \dfrac{1+i}{1-i} \right)}^{\dfrac{n}{3}}}=1$ by multiplying and dividing by the conjugate. The conjugate of this equation will be $\left( \dfrac{1+i}{1+i} \right)$.
$\begin{align}
  & {{\left( \dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i} \right)}^{\dfrac{n}{3}}}=1 \\
 & {{\left( \dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1+i \right)\left( 1-i \right)} \right)}^{\dfrac{n}{3}}}=1 \\
 & {{\left( \dfrac{1+{{i}^{2}}+2i}{1-{{i}^{2}}} \right)}^{\dfrac{n}{3}}}=1
\end{align}$
We will now simplify the expression by substituting the value \[{{i}^{2}}=-1\],
$\begin{align}
  & {{\left( \dfrac{1-1+2i}{1-\left( -1 \right)} \right)}^{\dfrac{n}{3}}}=1 \\
 & {{\left( \dfrac{2i}{2} \right)}^{\dfrac{n}{3}}}=1 \\
 & {{\left( i \right)}^{\dfrac{n}{3}}}=1 \\
 & {{\left( i \right)}^{\dfrac{n}{3}}}={{i}^{4}} \\
 & \dfrac{n}{3}=4 \\
 & n=12
\end{align}$
The least value we derived for $m$ and $n$ is $m=8$ and $n=12$.
We will now find the greatest common divisor of both the integers $m$ and $n$. We have to find the greatest integer which divides both the integers $m=8$ and $n=12$.
$4$ is the highest integer which divides both $m$ and $n$, hence the greatest common divisor will be $4$.

Therefore, the greatest common divisor of the least values of $m$ and $n$ is $4$.

Note: To rationalize, we multiply and divide the equation by its conjugate. Conjugates are the same expressions of a binomial but with different arithmetic operators. For example: The conjugate of a binomial $a+b$ is $a-b$.