Why is the argument and not ?

Because the real part is negative ( ) and the imaginary part is positive ( ), placing in the second quadrant where principal argument lies in , given by .

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Let

z = 1 + i

and

z_1 = \frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}

. Then

\frac{12}{\pi}\text{arg}(z_1)

is equal to

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Step-by-step solutionView

Correct answer

The value of $\frac{12}{\pi}\text{arg}(z_1)$ is $9$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$z = 1 + i$ and $z_1 = \frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$ .

02goal

Evaluate $\frac{12}{\pi}\text{arg}(z_1)$ using the principal argument of $z_1$ .

03approach

Substitute $z = 1 + i$ , $\bar{z} = 1 - i$ , and $1/z = \frac{1-i}{2}$ directly into the numerator and denominator of $z_1$ , simplify to the standard form $x + iy$ , and find $\text{arg}(z_1)$ .

04execute

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

05execute

Combine numerator and denominator to simplify $z_1$ : $z_1 = \frac{2 + i}{-\frac{1 + 3i}{2}} = -\frac{2(2 + i)}{1 + 3i} = -\frac{2(2 + i)(1 - 3i)}{1^2 + 3^2} = -\frac{2(2 - 6i + i + 3)}{10} = -\frac{2(5 - 5i)}{10} = -(1 - i) = -1 + i.$

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. $\text{arg}(z_1) = \pi - \tan^{-1}\left(\left|\frac{1}{-1}\right|\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.$

07execute

Compute the final scaled quantity: $\frac{12}{\pi}\text{arg}(z_1) = \frac{12}{\pi} \times \frac{3\pi}{4} = 9.$

✓verify

Check that $z_1 = -1 + i$ : $|z_1| = \sqrt{2}$ . $|1 + i\bar{z}| = |2 + i| = \sqrt{5}$ . Denominator is $|-\frac{1+3i}{2}| = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}$ . 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 $\bar{z} = 1 - i$ , simplify the numerator $1 + i\bar{z}$ and denominator $\bar{z}(1-z) + \frac{1}{z}$ separately in Cartesian form, then divide.
What does the numerator $1 + i\bar{z}$ evaluate to?
$2 + i$
What does the denominator $\bar{z}(1-z) + \frac{1}{z}$ evaluate to?
$\frac{-1 - 3i}{2}$
What are the simplified Cartesian form and argument of $z_1 = \frac{2(2+i)}{-1-3i}$ ?
$z_1 = -1 + i$ , so $\text{arg}(z_1) = \frac{3\pi}{4}$
Using $\text{arg}(z_1) = \frac{3\pi}{4}$ , what is the value of $\frac{12}{\pi}\text{arg}(z_1)$ ?
$9$

✓ Source and academic review

Question type: Numerical
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

3\pi/4

and not

-\pi/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 $(\pi/2, \pi]$ , given by $\pi - \tan^{-1}(1) = 3\pi/4$ .

Answer

The value of $\frac{12}{\pi}\text{arg}(z_1)$ is $9$ .

Why each option works or fails

Step-by-step solution

given: $z = 1 + i$ and $z_1 = \frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$ .
goal: Evaluate $\frac{12}{\pi}\text{arg}(z_1)$ using the principal argument of $z_1$ .
approach: Substitute $z = 1 + i$ , $\bar{z} = 1 - i$ , and $1/z = \frac{1-i}{2}$ directly into the numerator and denominator of $z_1$ , simplify to the standard form $x + iy$ , and find $\text{arg}(z_1)$ .
execute: Evaluate the numerator: $1 + i\bar{z} = 1 + i(1 - i) = 1 + i - i^2 = 1 + i + 1 = 2 + i$ . Evaluate the denominator: $\bar{z}(1 - z) + \frac{1}{z} = (1 - i)(1 - (1 + i)) + \frac{1 - i}{2} = (1 - i)(-i) + \frac{1 - i}{2} = (-i - 1) + \frac{1 - i}{2} = -\frac{1}{2} - \frac{3i}{2} = -\frac{1 + 3i}{2}$ .
execute: Combine numerator and denominator to simplify $z_1$ : $z_1 = \frac{2 + i}{-\frac{1 + 3i}{2}} = -\frac{2(2 + i)}{1 + 3i} = -\frac{2(2 + i)(1 - 3i)}{1^2 + 3^2} = -\frac{2(2 - 6i + i + 3)}{10} = -\frac{2(5 - 5i)}{10} = -(1 - i) = -1 + i.$
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. $\text{arg}(z_1) = \pi - \tan^{-1}\left(\left|\frac{1}{-1}\right|\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.$
execute: Compute the final scaled quantity: $\frac{12}{\pi}\text{arg}(z_1) = \frac{12}{\pi} \times \frac{3\pi}{4} = 9.$
verify: Check that $z_1 = -1 + i$ : $|z_1| = \sqrt{2}$ . $|1 + i\bar{z}| = |2 + i| = \sqrt{5}$ . Denominator is $|-\frac{1+3i}{2}| = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}$ . Thus $|z_1| = \frac{\sqrt{5}}{\sqrt{5/2}} = \sqrt{2}$ , which matches $|-1 + i| = \sqrt{2}$ .

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

The value of $\frac{12}{\pi}\text{arg}(z_1)$ is $9$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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