Question

If m and n are the smallest positive integers satisfying the relation \[{{\left( 2cis\dfrac{\pi }{6} \right)}^{m}}={{\left( 4\cos \dfrac{\pi }{4} \right)}^{n}},\] then (m + n) has the value equal to:
(a) 36
(b) 96
(c) 72
(d) 60

Answer 1

Hint: To solve this question, first we will convert \[cis\theta \] into the terms of sine and cosine by using the formula \[cis\theta =\cos \theta +i\sin \theta .\] After doing this, we will get the terms like \[{{\left( \cos \theta +i\sin \theta \right)}^{t}}\] on both the sides of the equation. We will write it as \[{{\left( \cos \theta +i\sin \theta \right)}^{t}}=\cos t\theta +i\sin t\theta .\] Then we will compare the real and imaginary parts of the equation obtained. From here, we will get a relation between m and n. Then we will obtain another equation by putting \[{{\left( \cos \theta +i\sin \theta \right)}^{t}}={{\left( {{e}^{i\theta }} \right)}^{t}}\] in the equation obtained earlier. When we will solve both these equations, we will get the values of m and n in terms of k. When we will put k = 1, we will get the smallest value of (m + n).

Complete step-by-step answer:
While we will solve this question, we will get the values of m and n in terms of a constant. To get the minimum value, we will put the value of the constant as 1. In the above question, we have,
\[{{\left( 2cis\dfrac{\pi }{6} \right)}^{m}}={{\left( 4\cos \dfrac{\pi }{4} \right)}^{n}}.....\left( i \right)\]
In equation (i), we are going to use the identity as shown:
\[{{\left( ab \right)}^{n}}={{a}^{n}}\times {{b}^{n}}\]
Thus, applying the identity in equation (i), we get the following result,
\[{{2}^{m}}{{\left( cis\dfrac{\pi }{6} \right)}^{m}}={{4}^{n}}{{\left( cis\dfrac{\pi }{4} \right)}^{n}}....\left( ii \right)\]
In the equation (ii), we are going to use another identity as shown below,
\[cis\theta =\cos \theta +i\sin \theta \]
Thus, after applying this identity, we get,
\[{{2}^{m}}{{\left( \cos \dfrac{\pi }{6}+i\sin \dfrac{\pi }{6} \right)}^{m}}={{4}^{n}}{{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{n}}.....\left( iii \right)\]
In the equation (iii), we are going to use the following identity,
\[{{\left( \cos \theta +i\sin \theta \right)}^{m}}=\cos m\theta +i\sin m\theta \]
Thus, we get,
\[{{2}^{m}}\left( \cos \dfrac{m\pi }{6}+i\sin \dfrac{m\pi }{6} \right)={{4}^{n}}\left( \cos \dfrac{n\pi }{4}+i\sin \dfrac{n\pi }{4} \right).....\left( iv \right)\]
Now, we will compare the real and imaginary parts in the above equation. We will compare the real part first.
\[{{2}^{m}}\cos \dfrac{m\pi }{6}={{4}^{n}}\cos \dfrac{n\pi }{4}\]
\[\Rightarrow \cos \dfrac{m\pi }{6}=\dfrac{{{4}^{n}}\cos \dfrac{n\pi }{4}}{{{2}^{m}}}....\left( v \right)\]
We know that, \[{{4}^{n}}={{2}^{2n}}.\] Thus, we get,
\[\Rightarrow \cos \dfrac{m\pi }{6}=\dfrac{{{2}^{2n}}\cos \dfrac{n\pi }{4}}{{{2}^{m}}}\]
Also, \[\dfrac{{{2}^{a}}}{{{2}^{b}}}={{2}^{a-b}}.\] Thus,
\[\Rightarrow \cos \dfrac{m\pi }{6}={{2}^{2n-m}}\cos \dfrac{n\pi }{4}.....\left( vi \right)\]
Similarly, now comparing the imaginary part, we get,
\[\sin \dfrac{m\pi }{6}={{2}^{2n-m}}\sin \dfrac{n\pi }{4}.....\left( vii \right)\]
Now, we will square (vi) and (vii) and then add them.
\[{{\left( \cos \dfrac{m\pi }{6} \right)}^{2}}+{{\left( \sin \dfrac{m\pi }{6} \right)}^{2}}={{\left( {{2}^{2n-m}}\cos \dfrac{n\pi }{4} \right)}^{2}}+{{\left( {{2}^{2n-m}}\sin \dfrac{n\pi }{4} \right)}^{2}}\]
\[\Rightarrow \left( {{\cos }^{2}}\dfrac{m\pi }{6} \right)+\left( {{\sin }^{2}}\dfrac{m\pi }{6} \right)={{\left( {{2}^{2n-m}} \right)}^{2}}\left( {{\cos }^{2}}\dfrac{n\pi }{4}+{{\sin }^{2}}\dfrac{n\pi }{4} \right)\]
In the above equation, we are going to use the following identity,
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]
Thus, we get,
\[1={{\left( {{2}^{2n-m}} \right)}^{2}}\left( 1 \right)\]
\[\Rightarrow {{2}^{2n-m}}=1\]
\[\Rightarrow {{2}^{2n-m}}={{2}^{0}}\]
As the bases are equal, powers will be equal. Thus, we get,
\[2n-m=0\]
\[\Rightarrow 2n=m.....\left( viii \right)\]
Now, we are going to use the following identity in (iii).
\[\cos \theta +i\sin \theta ={{e}^{i\theta }}\]
Thus, we get,
\[{{2}^{m}}{{\left( {{e}^{\dfrac{i\pi }{6}}} \right)}^{m}}={{4}^{n}}{{\left( {{e}^{\dfrac{i\pi }{4}}} \right)}^{n}}\]
\[\Rightarrow {{2}^{m}}{{e}^{\dfrac{im\pi }{6}}}={{4}^{n}}{{e}^{\dfrac{in\pi }{4}}}\]
\[\Rightarrow \dfrac{{{e}^{\dfrac{im\pi }{6}}}}{{{e}^{\dfrac{in\pi }{4}}}}=\dfrac{{{4}^{n}}}{{{2}^{m}}}\]
\[\Rightarrow {{e}^{i\left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)}}={{2}^{2n-m}}.....\left( ix \right)\]
Now, 2n = m. Thus, we get,
\[\Rightarrow {{e}^{i\left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)}}={{2}^{2n-2n}}\]
\[\Rightarrow {{e}^{i\left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)}}={{2}^{0}}\]
\[\Rightarrow {{e}^{i\left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)}}=1\]
Now, we know that \[{{e}^{i\theta }}=\cos \theta +i\sin \theta .\] Thus, we get,
\[\cos \left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)+i\sin \left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)=1\]
Now, comparing the real parts, we get,
\[\cos \left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)=\cos 2k\pi \]
\[\Rightarrow \dfrac{m\pi }{6}-\dfrac{n\pi }{4}=2k\pi \]
\[\Rightarrow \dfrac{m\pi }{6}-\dfrac{\left( \dfrac{m}{2} \right)\pi }{4}=2k\pi \]
\[\Rightarrow \dfrac{m}{6}-\dfrac{m}{8}=2k\]
\[\Rightarrow \dfrac{8m-6m}{48}=2k\]
\[\Rightarrow \dfrac{2m}{48}=2k\]
\[\Rightarrow \dfrac{m}{48}=2k\]
\[\Rightarrow m=48k\]
Thus, \[n=\dfrac{m}{2},\] therefore we get,
\[n=\dfrac{48k}{2}=24k\]
Now, to get the lowest value of m and n, we put k = 1. Thus, m = 48 and n = 24.
Therefore, m + n = 24 + 48 = 72.

So, the correct answer is “Option (c)”.

Note: We can also derive the relation between m and n as follows: We have,
\[{{2}^{m}}{{e}^{\dfrac{im\pi }{6}}}={{4}^{n}}{{e}^{\dfrac{in\pi }{4}}}\]
\[\Rightarrow {{e}^{i\left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)}}={{2}^{2n-m}}\]
\[\Rightarrow \cos \left( \dfrac{m\pi }{6}-\dfrac{n\pi }{4} \right)={{2}^{2n-m}}\]
Here, we can take only positive values of m and n. Also, we are going to apply the boundary condition. The maximum value of LHS is equal to the minimum value of RHS. The maximum value of the cos function is 1. Thus,
\[{{2}^{2n-m}}=1\]
\[\Rightarrow {{2}^{2n-m}}={{2}^{0}}\]
\[\Rightarrow 2n=m\]