Question

The sum of roots of the equation $x^8=1$ whose real part is positive is

• A. $1+\sqrt{2}$
• B. $1-\sqrt{2}$
• C. $0$
• D. $1$

Answer 1

The correct option is A $1+\sqrt{2}$

Given $x^8=1$
the roots can be $e^{i2k\pi /8},k=0,1,2,\cdots ,7$
Hence $x=e^{0\cdot i},e^{i\pi /4},e^{i\pi /2},e^{i3\pi /4},e^{i\pi },e^{i5\pi /4},e^{i3\pi /2},e^{i7\pi /4}$
So the roots with $+ive$ real part are
$e^{0\cdot i},e^{i\pi /4},e^{i7\pi /4}$
Hence required sum is
$e^{0\cdot i}+e^{i\pi /4}+e^{i7\pi /4}$$=\cos 0+\cos \pi /4+\cos \left(7\pi /4\right)=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=1+\sqrt{2}$