Question

If z = √(3)/2 + i/2(i + √-1), then
(1 + iz + z⁵ + iz⁸)⁹ is equal to:
(A) –1 (B) 1
(C) (–1 + 2i)⁹
(D) 0

Answer 1

Answer:

option c................

Answer 2

If z = √(3)/2 + i/2, (i = √-1), then  (1 + iz + z⁵ + iz⁸)⁹ is equal to -1

• z = cos( 30) + i sin (30)
• We know that  cos(x) + i sin(x) = $e^{ix}$ , by Euler's Formula.

Therefore,

• z = cos( 30) + i sin (30) =$cos(\frac{\pi }{6} )$ + i sin($\frac{\pi }{6}$)  = $e^{i\frac{\pi }{6} }$

Therefore,

• (1 + iz + z⁵ + iz⁸)⁹ = (1 + i$e^{i\frac{\pi }{6} }$  +$e^{i\frac{5\pi }{6} }$ + i$e^{\frac{i8\pi }{6} }$)^9.

Also,

• i = $e^{\frac{i\pi }{2} }$

• Let y = (1 + i$e^{i\frac{\pi }{6} }$  +$e^{i\frac{5\pi }{6} }$ + i$e^{\frac{i8\pi }{6} }$) = ( 1 +  $e^{\frac{i\pi }{2} }$.

• ( 1 +  $e^{\frac{i\pi }{2} }$.$e^{i\frac{\pi }{6} }$ + $e^{i\frac{5\pi }{6} }$ +  

Again applying Euler's formula,

• ( 1 + $e^{\frac{i2\pi }{3} }$ + $e^{i\frac{5\pi }{6} }$ + $e^{\frac{i11\pi }{6} }$)  =

            (1 + (cos($\frac{2\pi }{3}$) + i sin(

• y = (1 + (−1/2+i$\sqrt{3}$/2 + (−

• y = cos($\frac{\pi }{3}$) + isin(

Therefore , $Y^{9}$ = $e^{\frac{i9\pi }{3} } = e^{i3\pi }$ = cos(3$\pi$) + isin(3

Therefore, (1 + iz + z⁵ + iz⁸)⁹ is equal to -1