For a non-zero complex number \(z\), let \(\arg (z)\) denote the principal argument of \(z\), with \(-\pi<\arg (z) \leq \pi\). Let \(\omega\) be the cube root of unity for which \(0<\arg (\omega)<\pi\). Let \[ \alpha=\arg \left(\sum_{n=1}^{2025}(-\omega)^{n}\right) . \] Then the value of \(\frac{3 \alpha}{\pi}\) is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Complex Numbers

For a non-zero complex number

z

, let

ar g (z)

denote the principal argument of

z

, with

- π < ar g (z) \leq π

. Let

ω

be the cube root of unity for which

0 < ar g (ω) < π

. Let

α = ar g (n = 1 \sum 2025 (- ω)^{n}) .

Then the value of

\frac{3 α}{π}

is ____.

Answer: -2

The cube root of unity with

0 < ar g (ω) < π

ω = e^{2 π i /3} = cos \frac{2 π}{3} + i sin \frac{2 π}{3} = - \frac{1}{2} + i \frac{3}{2}

.

The sum is a geometric series:

n = 1 \sum 2025 (- ω)^{n} = (- ω) \frac{1 - ( - ω ) ^{2025}}{1 - ( - ω )}

Since

ω^{3} = 1

, we have

(- ω)^{3} = - 1

, so

(- ω)^{2025} = (- ω)^{3 \cdot 675} = (- 1)^{675} = - 1

.

Thus:

n = 1 \sum 2025 (- ω)^{n} = (- ω) \frac{1 - ( - 1 )}{1 + ω} = \frac{- 2 ω}{1 + ω}

Since

1 + ω + ω^{2} = 0

, we have

1 + ω = - ω^{2}

.

So:

\frac{- 2 ω}{1 + ω} = \frac{- 2 ω}{- ω ^{2}} = \frac{2}{ω} = 2 ω^{2}

Now

ω^{2} = e^{4 π i /3}

, so

ar g (ω^{2}) = \frac{4 π}{3} - 2 π = - \frac{2 π}{3}

(principal argument).

Therefore

ar g (2 ω^{2}) = - \frac{2 π}{3}

.

Thus

α = - \frac{2 π}{3}

, and:

\frac{3 α}{π} = \frac{3 \cdot ( - 2 π /3 )}{π} = - 2

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer