Question

If $m$ and $n$ are the smallest positive integers satisfying the relation ${\left(2Cis\frac{\pi }{6}\right)}^m={\left(4Cis\frac{\pi }{4}\right)}^n$, then $(m+n)$ has the value equal to

• A. $36$
• B. $96$
• C. $72$
• D. $60$

Answer 1

Answer: (A)

$36$

$Cis\left(x\right)=\cos x+i\sin x=e^{ix}$

$\therefore (2Cis\frac{\pi }{6}{)}^m=(2e^{i\pi /6}{)}^m$ and $(4Cis\frac{\pi }{4}{)}^n=(4e^{i\pi /4}{)}^n$

$\therefore (2e^{i\pi /6}{)}^m=(4e^{i\pi /4}{)}^n$

$2^m{e}^{-im\pi /6}=4^n{e}^{-in\pi /4}$

To eliminate $i$,

$m$ & $n$ should be a multiple of $LCM(6,4)$

$LCM(6,4)=12$

for smallest

Let $m=12n$ should be multiple of $12$

$\therefore 2^m.1=4^n$

$2^m=2^{2n}$

If $m=12n=6$ not a multiple of $12$

If $m=24n=12$

$\therefore m+n=24+12=36$