Complex Numbers
Complex Numbers
star_batch_jee_advanced_2025
Grade 11
Question:
Let $\alpha = \cos\frac{2\pi}{5} + i\sin\frac{2\pi}{5}$ and let $A_k = x + y\alpha^k + p\alpha^{2k} + w\alpha^{3k} + f\alpha^{4k}$ where $x, y, p, w, f$ are points on the circle $|z| = 1$, then $\frac{|A_0|^2 + |A_1|^2 + |A_2|^2 + |A_3|^2 + |A_4|^2}{5}$ is equal to ___________.
Step-by-Step Solution
Key Concept: The orthogonality relation $\sum_{k=0}^{4}\alpha^{jk}=5\delta_{j,0}$ for 5th roots of unity eliminates cross terms when squaring the sum.
Let $\alpha = e^{2\pi i/5}$, a primitive 5th root of unity. Since $x, y, p, w, f$ lie on $|z|=1$, we have $|x|=|y|=|p|=|w|=|f|=1$. Note that $A_k = x + y\alpha^k + p\alpha^{2k} + w\alpha^{3k} + f\alpha^{4k}$. Computing $\sum_{k=0}^{4}|A_k|^2 = \sum_{k=0}^{4}A_k\overline{A_k}$ and using orthogonality of 5th roots of unity: $\sum_{k=0}^{4}\alpha^{jk} = 0$ for $j \not\equiv 0 \pmod{5}$ and $=5$ for $j\equiv 0$. This yields $\sum_{k=0}^{4}|A_k|^2 = 5(|x|^2 + |y|^2 + |p|^2 + |w|^2 + |f|^2) = 5(1+1+1+1+1) = 25$. Therefore $\frac{\sum_{k=0}^{4}|A_k|^2}{5} = \frac{25}{5} = 5$.
Correct Answer: 5