If z={1+i{3}}{1-i{3}} where i = √-1 then what is the argument of z ?

Correct Answer is : 2π32π /332π

Correct Answer is : 2π32π /332π

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UPSC NDA I 2023 Mathematics Solved Paper

Question : 12 of 120

Marks: +1, -0

z=\frac{1+i\sqrt{3}}{1-i\sqrt{3}}

where i = √-1 then what is the argument of z ?

Solution:

Concept:

The argument of a complex number

z = a + ib

\arg(z) = \tan^{-1}\left(\frac{b}{a}\right)

, adjusted for the correct quadrant.

Explanation:

Given

z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}

Multiply numerator and denominator by the conjugate

(1 + i\sqrt{3})

z = \frac{(1 + i\sqrt{3})^2}{1 - (i\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1 + 3}

Simplify

(i\sqrt{3})^2 = -3

, so numerator becomes

1 - 3 + 2i\sqrt{3} = -2 + 2i\sqrt{3}

Thus

z = \frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}

Here

a = -\frac{1}{2}

(negative) and

b = \frac{\sqrt{3}}{2}

(positive), placing

z

in the second quadrant.

Compute the reference angle:

\tan^{-1}\left(\frac{b}{|a|}\right) = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

Therefore,

\arg(z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

Answer:

\frac{2\pi}{3}

(Option B).

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