Question

The integral solution of $(1-i{)}^x=2^x$ is:

• A. $x=1$
• B. $x=-1$
• C. $x=2$
• D. $x=0$

Answer 1

Answer: (D)

$x=0$

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

$⟹2^{\frac{x}{2}}{(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})}^x=2^x$

$⟹2^{\frac{x}{2}}{(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4})}^x=2^x$

$⟹(2{)}^{\frac{x}{2}}{e}^{-i\frac{x\pi }{4}}=2^x$
Thus comparing the powers gives us

$\frac{x}{2}=x$

$⟹2x=x$
$⟹x=0$.

Hence, option D is correct.