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Complex Numbers question

2025 · Shift 2 · Q29
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Complex Numbers question

2025 · Shift 2 · Q29

JEE AdvancedMathematicsComplex NumbersNumerical+4 / −1
For a non-zero complex number zzz, let arg⁡(z)\arg (z)arg(z) denote the principal argument of zzz, with −π-\pi−πα=arg⁡(∑n=12025(−ω)n)\alpha=\arg \left(\sum\limits_{n=1}^{2025}(-\omega)^n\right)α=arg(n=1∑2025​(−ω)n)ThenthevalueofThen the value ofThenthevalueof\frac{3 \alpha}{\pi}isisis\underline{\hspace{2cm}}$.
Numerical answer
View written solutionFree

Correct answer: -2

  1. We need to find
α=arg⁡(∑n=12025(−ω)n),\alpha=\arg\left(\sum_{n=1}^{2025}(-\omega)^n\right),α=arg(n=1∑2025​(−ω)n),

where ω\omegaω is the cube root of unity.

  1. For cube roots of unity,
ω3=1,ω≠1,\omega^3=1,\quad \omega\neq 1,ω3=1,ω=1,

and

1+ω+ω2=0.1+\omega+\omega^2=0.1+ω+ω2=0.

Also,

ω=e2πi/3=−12+32i.\omega=e^{2\pi i/3}=-\frac12+\frac{\sqrt3}{2}i.ω=e2πi/3=−21​+23​​i.

Hence

−ω=e2πi/3+πi=e5πi/3=e−iπ/3.-\omega=e^{2\pi i/3+\pi i}=e^{5\pi i/3}=e^{-i\pi/3}.−ω=e2πi/3+πi=e5πi/3=e−iπ/3.

So −ω-\omega−ω is a sixth root of unity.

  1. Let
S=∑n=12025(−ω)n.S=\sum_{n=1}^{2025}(-\omega)^n.S=n=1∑2025​(−ω)n.

This is a geometric progression with ratio r=−ωr=-\omegar=−ω.

Since

(−ω)3=(−1)3ω3=−1,(-\omega)^3=(-1)^3\omega^3=-1,(−ω)3=(−1)3ω3=−1,

we get

(−ω)6=1.(-\omega)^6=1.(−ω)6=1.

Therefore the powers repeat every 666 terms.

  1. Now compute how many complete cycles of 666 occur in 202520252025 terms:
2025=6⋅337+3.2025=6\cdot 337+3.2025=6⋅337+3.

So

S=337(∑n=16(−ω)n)+∑n=13(−ω)n.S=337\left(\sum_{n=1}^6(-\omega)^n\right)+\sum_{n=1}^3(-\omega)^n.S=337(n=1∑6​(−ω)n)+n=1∑3​(−ω)n.
  1. Since (−ω)6=1(-\omega)^6=1(−ω)6=1 and −ω≠1-\omega\neq 1−ω=1,
∑n=16(−ω)n=0.\sum_{n=1}^6(-\omega)^n=0.n=1∑6​(−ω)n=0.

Hence

S=∑n=13(−ω)n.S=\sum_{n=1}^3(-\omega)^n.S=n=1∑3​(−ω)n.
  1. Let r=−ωr=-\omegar=−ω. Then
S=r+r2+r3.S=r+r^2+r^3.S=r+r2+r3.

But r3=−1r^3=-1r3=−1, so

S=r+r2−1.S=r+r^2-1.S=r+r2−1.

Now,

r=−ω,r2=ω2.r=-\omega,\qquad r^2=\omega^2.r=−ω,r2=ω2.

Thus

S=−ω+ω2−1.S=-\omega+\omega^2-1.S=−ω+ω2−1.

Using

ω=−12+32i,ω2=−12−32i,\omega=-\frac12+\frac{\sqrt3}{2}i, \qquad \omega^2=-\frac12-\frac{\sqrt3}{2}i,ω=−21​+23​​i,ω2=−21​−23​​i,

we get

−ω=12−32i.-\omega=\frac12-\frac{\sqrt3}{2}i.−ω=21​−23​​i.

Therefore

S=(12−32i)+(−12−32i)−1=−1−3 i.S=\left(\frac12-\frac{\sqrt3}{2}i\right)+\left(-\frac12-\frac{\sqrt3}{2}i\right)-1 = -1-\sqrt3\,i.S=(21​−23​​i)+(−21​−23​​i)−1=−1−3​i.
  1. Now find its principal argument. The complex number
−1−3 i-1-\sqrt3\,i−1−3​i

lies in the third quadrant, with

tan⁡θ=−3−1=3.\tan \theta=\frac{-\sqrt3}{-1}=\sqrt3.tanθ=−1−3​​=3​.

Its standard angle is 4π3\frac{4\pi}{3}34π​, so the principal argument is

α=−2π3.\alpha=-\frac{2\pi}{3}.α=−32π​.
  1. Therefore,
3απ=3π(−2π3)=−2.\frac{3\alpha}{\pi}=\frac{3}{\pi}\left(-\frac{2\pi}{3}\right)=-2.π3α​=π3​(−32π​)=−2.

So the required integer is

−2.\boxed{-2}.−2​.
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