Let z = 1 + i and z 1 = 1 + i z z ( 1 - z ) + 1 z . Then 12 arg ( z 1 ) is equal to __________ . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
$Let z = 1 + i and z_{1} = \frac{1 + i \bar{z}}{\bar{z} (1 - z) + \frac{1}{z}} . Then \frac{12}{π} \arg (z_{1}) is equal to$ __________ . [2023]

Ans.
(9)

We have $z = 1 + i$ and $z_{1} = \frac{1 + i \bar{z}}{\bar{z} (1 - z) + \frac{1}{z}}$

$= \frac{1 + i (1 - i)}{(1 - i) [1 - (1 + i)] + \frac{1}{1 + i}} = - 1 + i$

Now, $\arg (z_{1}) = π - \tan^{- 1} (1) = π - \frac{π}{4} = \frac{3 π}{4}$

$\frac{12}{π} \arg (z_{1}) = \frac{12}{π} \times \frac{3 π}{4} = 9$

Want Us to Email you About Special Offers & Updates?