Posted by Vasu Chaudhary 4 months, 1 week ago

- 1 answers

Related Questions:

Posted by Ritik Rathore 9 months, 3 weeks ago

- 1 answers

Posted by Pragya . 5 months ago

- 0 answers

Posted by Muskan Chhilar 3 months, 2 weeks ago

- 1 answers

Posted by Abhishek Chauhan 2 years ago

- 0 answers

Posted by B.Ushaasree B.Ushaasree 5 months, 3 weeks ago

- 0 answers

Posted by Sudhanshu Mishra 1 year, 1 month ago

- 2 answers

Posted by Ayush Agrawal 1 year, 5 months ago

- 1 answers

Posted by Anuj Kumar 11 months, 1 week ago

- 7 answers

Subscribe complete study pack and get unlimited access to selected subjects. Effective cost is only ₹ 12.5/- per subject per month. Based on the latest CBSE & NCERT syllabus.

Create papers in minutes

Print with your name & Logo

Download as PDF

3 Lakhs+ Questions

Solutions Included

Based on CBSE Blueprint

Best fit for Schools & Tutors

- Work from home with us
- Create questions or review them from home

No software required, no contract to sign. Simply apply as teacher, take eligibility test and start working with us. Required desktop or laptop with internet connection

Sia 🤖 4 months, 1 week ago

1. Let {tex}x + yi = \sqrt {1 - i} {/tex}

Squaring both sides, we get

x

^{2}- y^{2}+ 2xyi = 1 - iEquating the real and imaginary parts

x

^{2}- y^{2}= 1 and 2xy=-1... . (i){tex}\therefore \;xy = \frac{{ - 1}}{2}{/tex}

Using the identity

(x

^{2}+ y^{2})^{2}= (x^{2}- y^{2})^{2}+ 4x^{2}y^{2}{tex} = {\left( 1 \right)^2} + 44{\left( { - \frac{1}{2}} \right)^2}{/tex}

= 1 + 1

= 2

{tex}\therefore \;{x^2} + {y^2} = \sqrt 2 {/tex} . . . . (ii) [Neglecting (-) sign as x

^{2}+ y^{2}> 0]Solving (i) and (ii) we get

{tex}{x^2} = \frac{{\sqrt 2 + 1}}{2}{/tex} and {tex}y = \frac{{\sqrt 2 - 1}}{2}{/tex}

{tex}\therefore x = \pm \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} and {tex}y = \pm \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}

Since the sign of xy is negative.

{tex}\therefore {/tex} if {tex}x = \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} then {tex}y = - \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}

and if {tex}x = - \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} then {tex}y = \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}

{tex}\therefore \sqrt {1 - i} = \pm \left( {\sqrt {\frac{{\sqrt 2 + 1}}{2}} - \sqrt {\frac{{\sqrt 2 - 1}}{2}} i} \right){/tex}

0Thank You