JEE Main 2020 — Mathematics Algebra

$${(\frac{1+i}{1-i})}^{\frac{m}{2}}={(\frac{1+i}{i-1})}^{\frac{n}{3}}=1$$(1−i1+i​)2m​=(i−11+i​)3n​=1

Rationalize both the terms

$${(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^{\frac{m}{2}}={(\frac{1+i}{i-1}\times \frac{-1-i}{-1-i})}^{\frac{n}{3}}=1$$(1−i1+i​×1+i1+i​)2m​=(i−11+i​×−1−i−1−i​)3n​=1

$${(\frac{1+{i}^{2}+2i}{1-{i}^{2}})}^{\frac{m}{2}}={(\frac{-1-{i}^{2}-2i}{{(-1)}^{2}-{i}^{2}})}^{\frac{n}{3}}=1$$(1−i21+i2+2i​)2m​=((−1)2−i2−1−i2−2i​)3n​=1

We know ${i}^{2}=-1$i2=−1

$${(\frac{2i}{2})}^{\frac{m}{2}}={(\frac{-2i}{2})}^{\frac{n}{3}}=1$$(22i​)2m​=(2−2i​)3n​=1

$${(i)}^{\frac{m}{2}}={(-i)}^{\frac{n}{3}}=1$$(i)2m​=(−i)3n​=1

For ${(i)}^{\frac{m}{2}}=1\Rightarrow \frac{m}{2}=4k\Rightarrow m=8k,k\in I$(i)2m​=1⇒2m​=4k⇒m=8k,k∈I

Since $m$m is natural number so the least value of $m$m is $8$8.

For ${(-i)}^{\frac{n}{3}}=1\Rightarrow \frac{n}{3}=4l\Rightarrow n=12l,l\in I$(−i)3n​=1⇒3n​=4l⇒n=12l,l∈I

Since $n$n is natural number so the least value of $n$n is $12$12.

So greatest common factor of $8$8 and $12$12 is $4$4.