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Argument of (1+√3i)¹⁰+(1−√3i)¹⁰ | Complex Numbers JEE

JEE Maths question with a full step-by-step solution.

Question

z = (1 + \sqrt{3}\,i)^{10} + (1 - \sqrt{3}\,i)^{10}

, then

\mathrm{Arg}(z)

may be

\dfrac{\pi}{2}

\pi

correct

\dfrac{\pi}{4}

Dnone of these

Solution

Step 1: Convert

1 + \sqrt 3\,i

to polar form. Its modulus is

\sqrt{1^2 + (\sqrt 3)^2} = \sqrt{4} = 2

, and its argument is

\tan^{-1}\!\dfrac{\sqrt 3}{1} = \dfrac{\pi}{3}

. Thus:

1 + \sqrt 3\,i = 2e^{i\pi/3}, \qquad 1 - \sqrt 3\,i = 2e^{-i\pi/3}

Step 2: Apply De Moivre's theorem to the tenth powers:

(1 + \sqrt 3\,i)^{10} = 2^{10} e^{i10\pi/3}, \qquad (1 - \sqrt 3\,i)^{10} = 2^{10} e^{-i10\pi/3}

Step 3: Add them. Using

e^{i\phi} + e^{-i\phi} = 2\cos\phi

z = 2^{10}\big(e^{i10\pi/3} + e^{-i10\pi/3}\big) = 2^{10}\cdot 2\cos\frac{10\pi}{3} = 2^{11}\cos\frac{10\pi}{3}

Step 4: Reduce the angle modulo

2\pi

\frac{10\pi}{3} - 2\pi = \frac{10\pi - 6\pi}{3} = \frac{4\pi}{3}, \qquad \cos\frac{4\pi}{3} = -\frac{1}{2}

Step 5: Compute

z

z = 2^{11}\cdot\left(-\frac{1}{2}\right) = -2^{10} = -1024

Step 6:

z

is a negative real number, which lies on the negative real axis, so its argument is

\pi

. Correct answer: (2)

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