Complex Numbers
De Moivre's Theorem – High Powers of ω
Complex Numbers_PYQ
Grade 11
Question:
If $i = \sqrt{-1}$, then $4 + 5\!\left(-\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right)^{334} + 3\!\left(-\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right)^{365}$ is equal to
$1-i\sqrt{3}$
$-1+i\sqrt{3}$
$i\sqrt{3}$
$-i\sqrt{3}$
Step-by-Step Solution
Key Concept: Any power of $\omega$ reduces by taking the exponent $\pmod{3}$. Then use $\omega=-1/2+i\sqrt{3}/2$ and $\omega^2=-1/2-i\sqrt{3}/2$ directly.
**Step 1: Identify the base as ω**
$-\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} = e^{i2\pi/3} = \omega$ (primitive cube root of unity).
**Step 2: Reduce exponents mod 3**
$334 = 3\times111+1 \Rightarrow \omega^{334}=\omega$. $\quad 365=3\times121+2 \Rightarrow \omega^{365}=\omega^2$.
**Step 3: Evaluate**
$4+5\omega+3\omega^2 = 4+5\!\left(-\tfrac{1}{2}+\tfrac{i\sqrt{3}}{2}\right)+3\!\left(-\tfrac{1}{2}-\tfrac{i\sqrt{3}}{2}\right) = 4-\tfrac{5}{2}-\tfrac{3}{2}+i\sqrt{3}\!\left(\tfrac{5}{2}-\tfrac{3}{2}\right) = 0+i\sqrt{3} = i\sqrt{3}$.
Correct Answer: 3