Question

Find the value of ${\left(\frac{1+i}{\sqrt{2}}\right)}^8+{\left(\frac{1-i}{\sqrt{2}}\right)}^8$

Answer 1

$\to {\left(\frac{1+i}{\sqrt{2}}\right)}^8+{\left(\frac{1-i}{\sqrt{2}}\right)}^8$

Now $\frac{1+i}{\sqrt{2}}=\left(1.e^{i\pi /4}\right)=e^{i\pi /4}$

$\frac{1-i}{\sqrt{2}}=\left(e^{-i\pi /4}\right)$

$\therefore$ By De Moviers Theorem

${\left(\frac{1+i}{\sqrt{2}}\right)}^8={\left(e^{i\pi /4}\right)}^8=e^{i\times 2\pi }=e^{2\pi i}=1$

${\left(\frac{1-i}{\sqrt{2}}\right)}^8={\left(e^{-i\pi /4}\right)}^8=e^{-2\pi i}=+1$

$\therefore {\left(\frac{1+i}{\sqrt{2}}\right)}^8+{\left(\frac{1-i}{\sqrt{2}}\right)}^8=1+1=2$