ISI2019-MMA-21

Question

A function $f:\mathbb{R^2} \rightarrow \mathbb{R}$ is called degenerate on $x_i$, if $f(x_1,x_2)$ remains constant when $x_i$ varies $(i=1,2)$. Define

$$f(x_1,x_2) = \mid 2^{\pi _i/x_1} \mid ^{x_2} \text{ for } x_1 \neq 0$$,

where $i = \sqrt {-1}$. Then which of the following statements is true?

• A. $f$ is degenerate on both $x_1$ and $x_2$
• B. $f$ is degenerate on $x_1$ but not on $x_2$
• C. $f$ is degenerate on $x_2$ but not on $x_1$
• D. $f$ is neither degenerate on $x_1$ nor on $x_2$

Comments

Question is not clear. Please take another photo.

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This question might need update as picture is not clear

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@pratekag , @akash.dinkar12 Edited the equation. Rest of the question has been typed correctly 

Answer 1

Let $z = \left | 2^{\pi i/x_{1}} \right |$ , then $f(x_{1},x_{2})  = z^{x_{2}}$

Now I will try to prove that $z=1.$ This is a question of theory of Complex Numbers on euler representation of complex numbers.

A small recap, $e^{i\theta} = \cos \theta + i \sin \theta$ and $ \left | e^{ i \theta} \right | = \sqrt{\cos ^{2} \theta+ \sin ^{2} \theta} = 1$ because  if $ z_1 = a+ib$ then $|z_1|= \sqrt{a^2 +b ^2}$

Now coming back to the problem,

Since,  $e^{ln\;x}=x$, So, $2^{\pi i/x_{1}}= e^{ln\; 2^{\pi i/x_{1}}}= e^{\pi (\ln2) i/x_{1} } $

Now, $z = \left | 2^{\pi i/x_{1}} \right |=\left | e^{\pi (\ln2) i/x_{1} }  \right | = \sqrt{\cos ^{2} \theta+ \sin ^{2} \theta} = 1$ where,
$\theta = \pi (\ln2) /x_{1} $ and $x_1 \neq 0$

Substituting back in $f$,

$f(x_{1},x_{2})  = 1^{x_{2}} = 1$

Hence $f$ is a constant function which means it is degenerate on both $x_{1}, x_{2}$
Option $A$ is correct

Comments

Please edit the value of $\theta$ in paragraph $6$. It cannot be imaginary.

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Edited. It is not imaginary in this case.