The value of ((1+ sin (2 π/9)+i cos (2 π/9)/1+ sin (2 π)9-i cos (2 π)9)3 is

Question

The value of ((1+ sin (2 π/9)+i cos (2 π/9)/1+ sin (2 π)9-i cos (2 π)9)3 is (A) -(1/2)(1-i √3) (B) (1/2)(√3+i) (C) -(1/2)(√3-i) (D) (1/2)(1-i √3)

Tardigrade Admin · Accepted Answer

Let sin (2 π/9)+ i cos (2 π/9)= z ((1+ z /1+ barz))3=((1+ z /1+(1) z ))3= z 3 ⇒( i ( cos (2 π/9)- i sin (2 π/9)))3 =-i( cos (2 π/3)- i sin (2 π/3))=-i((-1/2)- i (√3/2)) ⇒ (-1/2)(√3-i) .

Q. The value of $(\frac{1 + s i n \frac{2 π}{9} + i c o s \frac{2 π}{9}}{1 + s i n \frac{2 π}{9} - i c o s \frac{2 π}{9}})^{3}$ is

$- \frac{1}{2} (1 - i 3)$

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

$- \frac{1}{2} (3 - i)$

$\frac{1}{2} (1 - i 3)$

Q. The value of (1+sin92π​−icos92π​1+sin92π​+icos92π​​)3 is

−21​(1−i3​)

21​(3​+i)

−21​(3​−i)

21​(1−i3​)

Let sin92π​+icos92π​=z (1+zˉ1+z​)3=(1+z1​1+z​)3=z3 ⇒(i(cos92π​−isin92π​))3 =−i(cos32π​−isin32π​)=−i(2−1​−i23​​) ⇒2−1​(3​−i).

Q. The value of $(\frac{1 + s i n \frac{2 π}{9} + i c o s \frac{2 π}{9}}{1 + s i n \frac{2 π}{9} - i c o s \frac{2 π}{9}})^{3}$ is

$- \frac{1}{2} (1 - i 3)$

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

$- \frac{1}{2} (3 - i)$

$\frac{1}{2} (1 - i 3)$