JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Argument of a Complex Number Expression

Q: Why is the argument 3π/4 and not -π/4?

Because the real part is negative (-1) and the imaginary part is positive (1), placing z_1 in the second quadrant where principal argument lies in (π/2, π], given by π - tan^(-1)(1) = 3π/4.

Let z = 1 + i and z_1 = (1+iz bar)/(z bar(1-z)+1/z). Then 12/πarg(z_1) is equal to

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Step-by-step solutionView

Correct answer

The value of $\frac{12}{π} arg (z_{1})$ is $9$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

$z = 1 + i$ and $z_{1} = \frac{1 + i \overset{ˉ}{z}}{\overset{ˉ}{z} (1 - z) + \frac{1}{z}}$ .

02Goal

Evaluate $\frac{12}{π} arg (z_{1})$ using the principal argument of $z_{1}$ .

03Approach

Substitute $z = 1 + i$ , $\overset{ˉ}{z} = 1 - i$ , and $1 / z = \frac{1 - i}{2}$ directly into the numerator and denominator of $z_{1}$ , simplify to the standard form $x + i y$ , and find $arg (z_{1})$ .

04Execute

Evaluate the numerator: $1 + i \overset{ˉ}{z} = 1 + i (1 - i) = 1 + i - i^{2} = 1 + i + 1 = 2 + i$ . Evaluate the denominator: $\overset{ˉ}{z} (1 - z) + \frac{1}{z} = (1 - i) (1 - (1 + i)) + \frac{1 - i}{2} = (1 - i) (- i) + \frac{1 - i}{2} = (- i - 1) + \frac{1 - i}{2} = - \frac{1}{2} - \frac{3 i}{2} = - \frac{1 + 3 i}{2}$ .

05Execute

Combine numerator and denominator to simplify $z_{1}$ : $z_{1} = \frac{2 + i}{- \frac{1 + 3 i}{2}} = - \frac{2 (2 + i)}{1 + 3 i} = - \frac{2 (2 + i) (1 - 3 i)}{1^{2} + 3^{2}} = - \frac{2 (2 - 6 i + i + 3)}{10} = - \frac{2 (5 - 5 i)}{10} = - (1 - i) = - 1 + i .$

06Execute

Find the principal argument of $z_{1} = - 1 + i$ : Since $x = - 1 < 0$ and $y = 1 > 0$ , $z_{1}$ lies in the second quadrant. $arg (z_{1}) = π - \tan^{- 1} (∣ \frac{1}{- 1} ∣) = π - \frac{π}{4} = \frac{3 π}{4} .$

07Execute

Compute the final scaled quantity: $\frac{12}{π} arg (z_{1}) = \frac{12}{π} \times \frac{3 π}{4} = 9.$

✓Verify

Check that $z_{1} = - 1 + i$ : $∣ z_{1} ∣ = \sqrt{2}$ . $∣ 1 + i \overset{ˉ}{z} ∣ = ∣ 2 + i ∣ = \sqrt{5}$ . Denominator is $∣ - \frac{1 + 3 i}{2} ∣ = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}$ . Thus $∣ z_{1} ∣ = \frac{\sqrt{5}}{\sqrt{5 / 2}} = \sqrt{2}$ , which matches $∣ - 1 + i ∣ = \sqrt{2}$ .

Hints that build this answer step by step

What is the most direct way to evaluate $z_{1}$ given $z = 1 + i$ ?
Compute $\overset{ˉ}{z} = 1 - i$ , simplify the numerator $1 + i \overset{ˉ}{z}$ and denominator $\overset{ˉ}{z} (1 - z) + \frac{1}{z}$ separately in Cartesian form, then divide.
What does the numerator $1 + i \overset{ˉ}{z}$ evaluate to?
$2 + i$
What does the denominator $\overset{ˉ}{z} (1 - z) + \frac{1}{z}$ evaluate to?
$\frac{- 1 - 3 i}{2}$
What are the simplified Cartesian form and argument of $z_{1} = \frac{2 (2 + i)}{- 1 - 3 i}$ ?
$z_{1} = - 1 + i$ , so $arg (z_{1}) = \frac{3 π}{4}$
Using $arg (z_{1}) = \frac{3 π}{4}$ , what is the value of $\frac{12}{π} arg (z_{1})$ ?
$9$

✓ Source and academic review

Question type: Numerical
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

3 π / 4

and not

- π / 4

Because the real part is negative ( $- 1$ ) and the imaginary part is positive ( $1$ ), placing $z_{1}$ in the second quadrant where principal argument lies in $(π / 2, π]$ , given by $π - \tan^{- 1} (1) = 3 π / 4$ .

Answer

The value of $\frac{12}{π} arg (z_{1})$ is $9$ .

Why each option works or fails

Step-by-step solution

Given: $z = 1 + i$ and $z_{1} = \frac{1 + i \overset{ˉ}{z}}{\overset{ˉ}{z} (1 - z) + \frac{1}{z}}$ .
Goal: Evaluate $\frac{12}{π} arg (z_{1})$ using the principal argument of $z_{1}$ .
Approach: Substitute $z = 1 + i$ , $\overset{ˉ}{z} = 1 - i$ , and $1 / z = \frac{1 - i}{2}$ directly into the numerator and denominator of $z_{1}$ , simplify to the standard form $x + i y$ , and find $arg (z_{1})$ .
Execute: Evaluate the numerator: $1 + i \overset{ˉ}{z} = 1 + i (1 - i) = 1 + i - i^{2} = 1 + i + 1 = 2 + i$ . Evaluate the denominator: $\overset{ˉ}{z} (1 - z) + \frac{1}{z} = (1 - i) (1 - (1 + i)) + \frac{1 - i}{2} = (1 - i) (- i) + \frac{1 - i}{2} = (- i - 1) + \frac{1 - i}{2} = - \frac{1}{2} - \frac{3 i}{2} = - \frac{1 + 3 i}{2}$ .
Execute: Combine numerator and denominator to simplify $z_{1}$ : $z_{1} = \frac{2 + i}{- \frac{1 + 3 i}{2}} = - \frac{2 (2 + i)}{1 + 3 i} = - \frac{2 (2 + i) (1 - 3 i)}{1^{2} + 3^{2}} = - \frac{2 (2 - 6 i + i + 3)}{10} = - \frac{2 (5 - 5 i)}{10} = - (1 - i) = - 1 + i .$
Execute: Find the principal argument of $z_{1} = - 1 + i$ : Since $x = - 1 < 0$ and $y = 1 > 0$ , $z_{1}$ lies in the second quadrant. $arg (z_{1}) = π - \tan^{- 1} (∣ \frac{1}{- 1} ∣) = π - \frac{π}{4} = \frac{3 π}{4} .$
Execute: Compute the final scaled quantity: $\frac{12}{π} arg (z_{1}) = \frac{12}{π} \times \frac{3 π}{4} = 9.$
Verify: Check that $z_{1} = - 1 + i$ : $∣ z_{1} ∣ = \sqrt{2}$ . $∣ 1 + i \overset{ˉ}{z} ∣ = ∣ 2 + i ∣ = \sqrt{5}$ . Denominator is $∣ - \frac{1 + 3 i}{2} ∣ = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}$ . Thus $∣ z_{1} ∣ = \frac{\sqrt{5}}{\sqrt{5 / 2}} = \sqrt{2}$ , which matches $∣ - 1 + i ∣ = \sqrt{2}$ .

Argument of a Complex Number Expression

What feels right?

The value of $\frac{12}{π} arg (z_{1})$ is $9$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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