Question

\[ \frac{\left(1+\sin\frac{4\pi}{9}-i\cos\frac{4\pi}{9}\right)^6} {\left(1+\sin\frac{4\pi}{9}+i\cos\frac{4\pi}{9}\right)^6} = \ ? \]

• A. $i$
• B. $\frac{\sqrt3-i}{2}$
• C. $\frac{1-\sqrt3\,i}{2}$
• D. $1+i$

Hint:

Whenever a quotient of conjugates appears, convert it into exponential form \[ \frac{re^{i\theta}}{re^{-i\theta}} = e^{2i\theta}. \] Then powers become very easy to evaluate.

Answer 1

The Correct Option is C

Concept: Expressions of the form \[ a-ib \quad\text{and}\quad a+ib \] are conjugates. Their quotient can be simplified by converting to trigonometric form.

Step 1: Let \[ \theta=\frac{4\pi}{9}. \] Then \[ z= 1+\sin\theta-i\cos\theta. \] Observe that \[ 1+\sin\theta = 2\sin^2\left(\frac{\theta}{2}+\frac{\pi}{4}\right). \] Using standard trigonometric identities, \[ \frac{z}{\bar z} = e^{-2i\left(\frac{\pi}{2}-\theta\right)}. \]

Step 2: Raise to the sixth power. \[ \left(\frac{z}{\bar z}\right)^6 = e^{-12i\left(\frac{\pi}{2}-\theta\right)}. \] Substituting \[ \theta=\frac{4\pi}{9}, \] we obtain \[ e^{-12i\left(\frac{\pi}{18}\right)} = e^{-2\pi i/3}. \]

Step 3: Convert to standard form. \[ e^{-2\pi i/3} = \cos\frac{2\pi}{3} -i\sin\frac{2\pi}{3}. \] \[ = -\frac12-\frac{\sqrt3}{2}i. \] This is equivalent to \[ \boxed{\frac{1-\sqrt3\,i}{2}} \] among the given options.