
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
163.8k+ views
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 $a-b$.
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 $a-b$.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
