Complex Numbers
Complex Fraction — De Moivre's Theorem
nta_pyq_2023_jan
Grade None
Question:
The value of $\left(\dfrac{1+\sin\dfrac{2\pi}{9}+i\cos\dfrac{2\pi}{9}}{1+\sin\dfrac{2\pi}{9}-i\cos\dfrac{2\pi}{9}}\right)^3$ is:
$\dfrac{-1}{2}(1-i\sqrt{3})$
$\dfrac{1}{2}(1-i\sqrt{3})$
$\dfrac{-1}{2}(\sqrt{3}-i)$
$\dfrac{1}{2}(\sqrt{3}+i)$
Step-by-Step Solution
Key Concept: Let $z=\sin(2\pi/9)+i\cos(2\pi/9)$. The fraction $=\frac{1+z}{1+\bar{z}}=z^? $... Use: numerator $=1+z$, denominator $=1+\bar{z}$. $\frac{1+z}{1+\bar{z}}=z$ (since $|z|=1$ and $1+z=z(z^{-1}+1)=z(\bar{z}+1)$). Actually $\left(\frac{1+z}{1+\bar z}\right)^3=z^3=i^3(\cos(2\pi/9)-i\sin(2\pi/9))^3$.
$\dfrac{-1}{2}(\sqrt{3}-i)$.
Correct Answer: 3