Question

Apply De Moivre's theorem to write (√3+i )^5 in the form a+ib , with a,b belongs to R

Answer 1

Let R = (√3 + i)⁵ , we should arrange it in standard form e.g., R = r(cosα +isinα)ⁿ
here , r = $\sqrt{a^2+b^2}$ here, R = a + ib
so, r = $\sqrt{\sqrt{3}^2+1^2}$ = 2
∴ R = 2⁵(√3/2 + 1/2i)⁵
= 2⁵(cos30° + isin30°)⁵

Now, Use De- moivre's theorem,
R = 2⁵(cos 3 × 30° + isin5 × 30°)
= 2⁵(cos150° + isin150°)
= 2⁵[cos(180-30) + isin (180°-30°)]
= 2⁵[ -cos30° + isin30° ]
= 2⁵[-√3/2 + 1/2i)
= 32( -√3/2 + 1/2i)
= -16√3 + 16i

Hence, answer is -16√3 + 16i