Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

if $(x+iy)^{2}=7+24i,$ then a value of $(7+\sqrt{-576})^{\frac{1}{2}}-(7-\sqrt{576})^{\frac{1}{2}}\; is$ :

Option 1)
$-6i$

Option 2)
$-3i$

Option 3)
$2i$

Option 4)
$6$

Answers (2)

best_answer

As we learnt in

Square Root of a Complex Number -

$\sqrt{z}=a+ib \ where\ z=x+iy \ is \ calcuated \ by\ equating\ real\ and \ imaginary \ parts \ of\ : \ x+iy=(a+ib)^2$

-

We have to find $(7+24i)^{1/2}-(7-24i)^{1/2}$

Now $(x+iy)^{2}=7+24i$

$x+iy=\sqrt{7+24i}$

and $x-iy=\sqrt{7-24i}$ (Put - i at the place of i)

$\therefore\ \ i2y=\sqrt{7+24i}-\sqrt{7-24i}$

Now x² - y² + i 2xy = 7 + 24 i

$\therefore\ \ x^{2}-y^{2}=7, \ \; xy = 12$

Solve for x & y, it is

$\therefore\ x=\pm4\ \; &\ \; y = \pm 3$

$\therefore\ \; i2y=\pm 6i$

$\therefore\ \; -6i$

Correct option is 1.

Option 1)

$-6i$

This is the correct option.

Option 2)

$-3i$

This is an incorrect option.

Option 3)

$2i$

This is an incorrect option.

Option 4)

$6$

This is an incorrect option.

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Andhra Pradesh cut-offs drop for CSE, ECE, other BTech programmes this year; closing rank

anam-khan

JEE Main 2025 Live: Exam city slip soon at jeemain.nta.nic.in; admit card date, cut-off updates

anam-khan

BTech Admissions 2025: GUJCET registration ends on December 31; JEE Main, AEEE, WBJEE, SRMJEEE exam date

Latest Question