Why is the argument of equal to instead of or ?

Write in standard form as . The real part is negative ( ) and the imaginary part is positive ( ), which places the number in the second quadrant. The argument is .

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

The complex number

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

is equal to :

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

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

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

\cos \frac{\pi}{12} - i \sin \frac{\pi}{12}

Step-by-step solutionView

Correct answer

Write $i-1$ in polar form as $\sqrt{2} e^{i\frac{3\pi}{4}}$ , then subtract the argument of the denominator $\frac{\pi}{3}$ to obtain $\sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$ .

Option analysis

Why each option works or fails

A ·

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

Factoring an extra factor of $i$ or confusing polar form conventions leads to an unsimplified product with an incorrect sign. Express the complex quotient directly as $r(\cos \theta + i \sin \theta)$ without introducing an extra factor of $i$ outside.

B ·

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

Taking $\arg(i-1)$ as $\frac{\pi}{4}$ instead of $\frac{3\pi}{4}$ gives the argument difference $\frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$ or miscalculating $\frac{3\pi}{4} - \frac{\pi}{3}$ as $\frac{\pi}{12}$ . Locate $-1 + i$ in the second quadrant, which has argument $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ , so $\frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$ .

C ·

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

This is the correct option. The modulus is $|i-1| / 1 = \sqrt{2}$ and the argument is $\arg(i-1) - \frac{\pi}{3} = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$ .

D ·

\cos \frac{\pi}{12} - i \sin \frac{\pi}{12}

Forgetting the modulus $\sqrt{2}$ entirely and subtracting the arguments in the wrong order or with incorrect signs. Ensure the numerator's modulus $|-1+i| = \sqrt{2}$ is factored in, and subtract the denominator's angle from the numerator's angle.

Reviewed route

Solution

StepWorking

01given

We are given the complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ .

02goal

Express $z$ in polar form $r(\cos \theta + i \sin \theta)$ .

03approach

Convert the numerator $i - 1 = -1 + i$ into Euler's form $r_1 e^{i\theta_1}$ , observe the denominator is already in Euler's form $e^{i\pi/3}$ , and use the quotient rule $\frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$ .

04execute

For the numerator: $-1 + i$ has modulus $|-1 + i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ . The point $(-1, 1)$ lies in the second quadrant, so its principal argument is $\arg(-1+i) = \pi - \tan^{-1}(1) = \frac{3\pi}{4}$ . Thus, $-1 + i = \sqrt{2} e^{i \frac{3\pi}{4}}$ .

05execute

The denominator is $\cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \frac{\pi}{3}}$ . Dividing gives $z = \frac{\sqrt{2} e^{i \frac{3\pi}{4}}}{e^{i \frac{\pi}{3}}} = \sqrt{2} e^{i\left(\frac{3\pi}{4} - \frac{\pi}{3}\right)} = \sqrt{2} e^{i \frac{5\pi}{12}} = \sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$ .

✓verify

Check quadrants: $-1+i$ has angle $135^\circ$ , denominator has angle $60^\circ$ . The difference is $135^\circ - 60^\circ = 75^\circ = \frac{5\pi}{12}$ . Modulus is $\sqrt{2}/1 = \sqrt{2}$ . Matches Option (2).

Hints that build this answer step by step

What is the polar form of the numerator, $-1 + i$ ?
$\sqrt{2} e^{i\frac{3\pi}{4}}$
Dividing $\sqrt{2} e^{i\frac{3\pi}{4}}$ by $e^{i\frac{\pi}{3}}$ , what is the resulting argument?
$\frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the argument of

i-1

equal to

\frac{3\pi}{4}

instead of

-\frac{\pi}{4}

\frac{\pi}{4}

Write $i - 1$ in standard form as $-1 + i$ . The real part is negative ( $-1$ ) and the imaginary part is positive ( $+1$ ), which places the number in the second quadrant. The argument is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ .

Answer

Write $i-1$ in polar form as $\sqrt{2} e^{i\frac{3\pi}{4}}$ , then subtract the argument of the denominator $\frac{\pi}{3}$ to obtain $\sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$ .

Why each option works or fails

A: $\sqrt{2} i \left(\cos \frac{5\pi}{12} - i \sin \frac{5\pi}{12}\right)$ - Factoring an extra factor of $i$ or confusing polar form conventions leads to an unsimplified product with an incorrect sign. Express the complex quotient directly as $r(\cos \theta + i \sin \theta)$ without introducing an extra factor of $i$ outside.
B: $\sqrt{2}\left(\cos \frac{\pi}{12} + i \sin \frac{\pi}{12}\right)$ - Taking $\arg(i-1)$ as $\frac{\pi}{4}$ instead of $\frac{3\pi}{4}$ gives the argument difference $\frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$ or miscalculating $\frac{3\pi}{4} - \frac{\pi}{3}$ as $\frac{\pi}{12}$ . Locate $-1 + i$ in the second quadrant, which has argument $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ , so $\frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$ .
C · correct: $\sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$ - This is the correct option. The modulus is $|i-1| / 1 = \sqrt{2}$ and the argument is $\arg(i-1) - \frac{\pi}{3} = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$ .
D: $\cos \frac{\pi}{12} - i \sin \frac{\pi}{12}$ - Forgetting the modulus $\sqrt{2}$ entirely and subtracting the arguments in the wrong order or with incorrect signs. Ensure the numerator's modulus $|-1+i| = \sqrt{2}$ is factored in, and subtract the denominator's angle from the numerator's angle.

Step-by-step solution

given: We are given the complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ .
goal: Express $z$ in polar form $r(\cos \theta + i \sin \theta)$ .
approach: Convert the numerator $i - 1 = -1 + i$ into Euler's form $r_1 e^{i\theta_1}$ , observe the denominator is already in Euler's form $e^{i\pi/3}$ , and use the quotient rule $\frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$ .
execute: For the numerator: $-1 + i$ has modulus $|-1 + i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ . The point $(-1, 1)$ lies in the second quadrant, so its principal argument is $\arg(-1+i) = \pi - \tan^{-1}(1) = \frac{3\pi}{4}$ . Thus, $-1 + i = \sqrt{2} e^{i \frac{3\pi}{4}}$ .
execute: The denominator is $\cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \frac{\pi}{3}}$ . Dividing gives $z = \frac{\sqrt{2} e^{i \frac{3\pi}{4}}}{e^{i \frac{\pi}{3}}} = \sqrt{2} e^{i\left(\frac{3\pi}{4} - \frac{\pi}{3}\right)} = \sqrt{2} e^{i \frac{5\pi}{12}} = \sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$ .
verify: Check quadrants: $-1+i$ has angle $135^\circ$ , denominator has angle $60^\circ$ . The difference is $135^\circ - 60^\circ = 75^\circ = \frac{5\pi}{12}$ . Modulus is $\sqrt{2}/1 = \sqrt{2}$ . Matches Option (2).

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

Write $i-1$ in polar form as $\sqrt{2} e^{i\frac{3\pi}{4}}$ , then subtract the argument of the denominator $\frac{\pi}{3}$ to obtain $\sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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