Question

(1+i√2)^8+(1-i√2)^8​

Answer 1

Answer:

Step-by-step explanation:

easy $(\frac{1}{\sqrt{2} } +\frac{i}{\sqrt{2} })^8+(\frac{1}{\sqrt{2} } -\frac{i}{\sqrt{2} })^8$

so we have to convert it into polar form ,

$[{cos(\frac{\pi }{4}) +isin(\frac{\pi }{4})]^8+[cos(\frac{\pi }{4} ) -isin(\frac{\pi }{4} )]^8$

In accordance with Euler's rule,

$(e^i^\frac{\pi }{4} )^8+(e^-^i^\frac{\pi }{4})^8$

$(e^i^2^\pi )+(e^-^i^2^\pi)$

again by polar form,

$(cos2\pi +isin2\pi )+(cos2\pi -isin2\pi )$

$=2cos2\pi$

$=2(1)$

=2