Question

Find the integral solution of the equation
${(1-i)}^n=2^n$

Answer 1

Suppose an integer $n$ satisfies the given equation Then ${(1-i)}^n=2^n$ We know that if two complex numbers are equal then their moduli are also equal. Hence taking moduli of both sides of $\left(1\right)$, we get $|{(1-i)}^n=2^n|$ or ${|1-i|}^n=2^n[\because 2^n>0,$ we have $\left|2^n\right||=2^n|$ or ${\left(\sqrt{2}\right)}^n=2^n$ Now $\left(2\right)$ will hold only when $n=0$. Hence the only integral solution of the given equation is $n=0$. Alternative: $1-i=\sqrt{2}(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4})$ Then given equation takes the form ${\left(\sqrt{2}\right)}^2{\left[\cos \right(-\pi /4)+i\sin (-\pi /4\left)\right]}^n=2^n$ or $2^{n/2}[\cos \frac{n\pi }{4}-i\sin \frac{n\pi }{4}]=2^n$ Equating real ad imaginary parts, we get $\cos \left(n\pi /4\right)=2^{n/2}$ and $-\sin \left(n\pi /4\right)=0$. These are satisfied only for $n=0$. Hence $n=0$ is the only solution.