JEE MainMathematics

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Converting a Quotient of Complex Numbers into Polar Form

Q: Why is the argument of i-1 equal to 3π/4 instead of -π/4 or π/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 π - π/4 = 3π/4.

The complex number z = (i-1)/(cos π/3 + i sin π/3) is equal to:

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A√(2) i (cos 5π/12 - i sin 5π/12)B√(2)(cos π/12 + i sin π/12)C√(2)(cos 5π/12 + i sin 5π/12)Dcos π/12 - i sin π/12

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · √(2) i (cos 5π/12 - i sin 5π/12)

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 $r (\cos θ + i \sin θ)$ without introducing an extra factor of $i$ outside.

B · √(2)(cos π/12 + i sin π/12)

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

C · √(2)(cos 5π/12 + i sin 5π/12)

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

D · cos π/12 - i sin π/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 $z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ .

02Goal

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

03Approach

Convert the numerator $i - 1 = - 1 + i$ into Euler's form $r_{1} e^{i θ_{1}}$ , observe the denominator is already in Euler's form $e^{i π / 3}$ , and use the quotient rule $\frac{r_{1} e^{i θ_{1}}}{r_{2} e^{i θ_{2}}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})}$ .

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 $\arg (- 1 + i) = π - \tan^{- 1} (1) = \frac{3 π}{4}$ . Thus, $- 1 + i = \sqrt{2} e^{i \frac{3 π}{4}}$ .

05Execute

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

✓Verify

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

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the argument of

i - 1

equal to

\frac{3 π}{4}

instead of

- \frac{π}{4}

\frac{π}{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 $π - \frac{π}{4} = \frac{3 π}{4}$ .

Answer

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

Why each option works or fails

A: √(2) i (cos 5π/12 - i sin 5π/12) - 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 $r (\cos θ + i \sin θ)$ without introducing an extra factor of $i$ outside.
B: √(2)(cos π/12 + i sin π/12) - Taking $\arg (i - 1)$ as $\frac{π}{4}$ instead of $\frac{3 π}{4}$ gives the argument difference $\frac{π}{4} - \frac{π}{3} = - \frac{π}{12}$ or miscalculating $\frac{3 π}{4} - \frac{π}{3}$ as $\frac{π}{12}$ . Locate $- 1 + i$ in the second quadrant, which has argument $π - \frac{π}{4} = \frac{3 π}{4}$ , so $\frac{3 π}{4} - \frac{π}{3} = \frac{5 π}{12}$ .
C · correct: √(2)(cos 5π/12 + i sin 5π/12) - This is the correct option. The modulus is $∣ i - 1 ∣ / 1 = \sqrt{2}$ and the argument is $\arg (i - 1) - \frac{π}{3} = \frac{3 π}{4} - \frac{π}{3} = \frac{5 π}{12}$ .
D: cos π/12 - i sin π/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 $z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ .
Goal: Express $z$ in polar form $r (\cos θ + i \sin θ)$ .
Approach: Convert the numerator $\arg (i - 1) - \frac{π}{3} = \frac{3 π}{4} - \frac{π}{3} = \frac{5 π}{12}$ into Euler's form $r_{1} e^{i θ_{1}}$ , observe the denominator is already in Euler's form $e^{i π / 3}$ , and use the quotient rule $\frac{r_{1} e^{i θ_{1}}}{r_{2} e^{i θ_{2}}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})}$ .
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 $\arg (- 1 + i) = π - \tan^{- 1} (1) = \frac{3 π}{4}$ . Thus, $- 1 + i = \sqrt{2} e^{i \frac{3 π}{4}}$ .
Execute: The denominator is $\cos \frac{π}{3} + i \sin \frac{π}{3} = e^{i \frac{π}{3}}$ . Dividing gives $z = \frac{\sqrt{2} e^{i \frac{3 π}{4}}}{e^{i \frac{π}{3}}} = \sqrt{2} e^{i (\frac{3 π}{4} - \frac{π}{3})} = \sqrt{2} e^{i \frac{5 π}{12}} = \sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$ .
Verify: Check quadrants: $- 1 + i$ has angle $135^{\circ}$ , denominator has angle $z = \frac{\sqrt{2} e^{i \frac{3 π}{4}}}{e^{i \frac{π}{3}}} = \sqrt{2} e^{i (\frac{3 π}{4} - \frac{π}{3})} = \sqrt{2} e^{i \frac{5 π}{12}} = \sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$ . The difference is $135^{\circ} - 60^{\circ} = 75^{\circ} = \frac{5 π}{12}$ . Modulus is $\sqrt{2} / 1 = \sqrt{2}$ . Matches Option C.

Converting a Quotient of Complex Numbers into Polar Form

What feels right?

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

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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