The complex number z = i-1 3 + i 3 is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2023Modulus & Argument

The complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ is equal to:

Answer

Correct answer:A

Standard Method

Given: $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$

Find: The equivalent polar form of $z$ and the correct option.

Using the trigonometric values,

\cos \frac{\pi}{3} = \frac{1}{2}, \qquad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}

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

Multiply numerator and denominator by the conjugate of the denominator:

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

Since

\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = 1

we get

z = (i-1)\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)

Expanding,

\begin{aligned} z &= i \cdot \frac{1}{2} - i \cdot \frac{\sqrt{3}}{2}i - \frac{1}{2} + \frac{\sqrt{3}}{2}i \\ &= \frac{\sqrt{3}-1}{2} + \frac{1+\sqrt{3}}{2}i \end{aligned}

Now compare with $r(\cos \theta + i \sin \theta)$ . The modulus is

r = \sqrt{\left(\frac{\sqrt{3}-1}{2}\right)^2 + \left(\frac{1+\sqrt{3}}{2}\right)^2} = \sqrt{2}

Hence,

r\cos \theta = \frac{\sqrt{3}-1}{2}, \qquad r\sin \theta = \frac{1+\sqrt{3}}{2}

So,

\cos \theta = \frac{\sqrt{3}-1}{2\sqrt{2}}, \qquad \sin \theta = \frac{1+\sqrt{3}}{2\sqrt{2}}

which gives

\theta = \frac{5\pi}{12}

Therefore,

z = \sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)

So the correct option is A.

The solution contains a contradictory label saying option D, but the worked expression and final polar form clearly match option A.

Use arguments directly

Given: $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$

Find: The polar form of $z$ .

Write each factor in polar form. First,

i - 1 = -1 + i

Its modulus is

\sqrt{(-1)^2 + 1^2} = \sqrt{2}

and its argument is in the second quadrant, so

\arg(i-1) = \frac{3\pi}{4}

Thus,

i-1 = \sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)

Also,

\cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \operatorname{cis}\frac{\pi}{3}

Therefore,

z = \sqrt{2} \, \operatorname{cis}\left(\frac{3\pi}{4} - \frac{\pi}{3}\right)

Now,

\frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi - 4\pi}{12} = \frac{5\pi}{12}

Hence,

z = \sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)

Therefore, the correct option is A.

Using the incorrect argument of $i-1$ . The number $i-1 = -1+i$ lies in the second quadrant, so its argument is $\frac{3\pi}{4}$ , not $\frac{\pi}{4}$ . Always check the quadrant from the signs of real and imaginary parts.
Choosing option D by copying the mislabeled line in the solution. The worked polar form has positive imaginary part, so it matches option A, not the option with a minus sign before $i \sin \frac{5\pi}{12}$ .
Rationalizing the denominator but expanding incorrectly. When multiplying $(i-1)\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$ , the term $-i^2\frac{\sqrt{3}}{2}$ becomes positive because $i^2=-1$ . Keep track of signs carefully.

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