Question

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

Answer 1

Step1. Finding value of $m$

${\left[\frac{1+i}{1-i}\right]}^{\frac{m}{2}}={\left[\frac{1+i}{i-1}\right]}^{\frac{n}{3}}=1$

$$\begin{gathered} \Rightarrow {\left[\frac{1+i}{1-i}\right]}^{\frac{m}{2}}=1 \\ \Rightarrow {[\frac{1+i}{1-i}\times \frac{1+i}{1+i}]}^{\frac{m}{2}}=1 \\ \Rightarrow {\left[\frac{{(1+i)}^2}{1^2-i^2}\right]}^{\frac{m}{2}}=1 \\ \Rightarrow {\left[\frac{1-1+2i}{2}\right]}^{\frac{m}{2}}=1 \\ \Rightarrow {\left[\frac{2i}{2}\right]}^{\frac{m}{2}}=1\mathbf{[}{\mathit{i}}^{\mathbf{2}}\mathbf{=}-1] \\ \Rightarrow i^{\frac{m}{2}}=i^4\mathbf{}\mathbf{[}{\mathit{i}}^{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{]} \\ \therefore \frac{m}{2}=4 \\ m=8 \end{gathered}$$

Step2. Finding value of $n$.

$$\begin{gathered} \Rightarrow {\left[\frac{1+i}{i-1}\right]}^{\frac{n}{3}}=1 \\ \Rightarrow {[\frac{1+i}{-1+i}\times \frac{-1-i}{-1-i}]}^{\frac{n}{3}}=1 \\ {\Rightarrow \left[\frac{-1-i-i+1}{{(-1)}^2-i^2}\right]}^{\frac{n}{3}}=1 \\ \Rightarrow {\left[\frac{-2i}{2}\right]}^{\frac{n}{3}}=1\mathbf{\{}{\mathit{i}}^{\mathbf{2}}\mathbf{=}\mathbf{1}\} \\ \Rightarrow {(-i)}^{\frac{n}{3}}={(-i)}^4\mathbf{\{}{\mathit{i}}^{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{\}} \\ \therefore \frac{n}{3}=4 \\ n=12 \end{gathered}$$

Step3.Finding the greatest common divisor.

Greatest common divisor can be calculated by.

Taking the product of both number i.e. $\begin{array}{rcl}m\times n& =& 12\times 8\\ & =& 96\end{array}$

Taking the L.C.M of both number i.e. L.C.M of $(12,8)$ is $24$

Now divide $96$ from $24$ we get $4$

Therefore, greatest common divisor of $m\&n$ is $4$