Question

Number of non-zero integral values of $x$ for which ${\left|1-i\right|}^x=2^x$ is

Answer 1

$|1-i{|}^x=a{2}^x$

$\Rightarrow |1-i{|}^x=(\sqrt{2}\times \sqrt{2}{)}^x$

$\Rightarrow {\left|\frac{1-i}{\sqrt{2}}\right|}^x=(\sqrt{2}{)}^x$

$\Rightarrow {\left|\cos \left(\frac{-\pi }{4}\right)\right|}^x=(\sqrt{2}{)}^x$

$\Rightarrow \cos \left(\frac{-x\pi }{4}\right)=(\sqrt{2}{)}^x$

$\Rightarrow \cos \left(\frac{x\pi }{4}\right)-i\sin \left(\frac{x\pi }{4}\right)=(\sqrt{2}{)}^x+i(0)$

$\Rightarrow \cos \left(\frac{x\pi }{4}\right)=(\sqrt{2}{)}^x;\sin \left(\frac{x\pi }{4}\right)=0x\in 4n$

if we put $x=4n$ in $\cos \left(\frac{x\pi }{4}\right)=(\sqrt{2}{)}^x$

$LHS\Rightarrow cos\left(\frac{x\pi }{4}\right)=\cos \left(\frac{4n\pi }{4}\right)$

$=\cos \left(n\pi \right)$

$=(-1{)}^n\ne (\sqrt{2}{)}^x$

$\therefore LHS\ne RHS$

$\therefore$ There is no ${}^{\prime }{x}^{\prime }$ that satisfies the coordinate

$\therefore$ There are ${}^{\prime }{o}^{\prime }$ non see integral solutions.