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Complex Numbers and Quadratic Equations

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ ∴ | x_{1} + y_{1} i | = | x_{2} + y_{2} i | \\ \Rightarrow \sqrt[]{x_{1}^{2} + y_{1}^{2}} = \sqrt[]{x_{2}^{2} + y_{2}^{2}} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} a n d y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} a n d y_{1} = \pm y_{2} \\ S o, z_{1} = x_{1} + y_{1} i a n d z_{2} = \pm x_{2} \pm y_{2} i \\ ∴ z_{1} \neq z_{2} \\ H e n c e, i t i s n o t n e c e s s a r y t h a t z_{1} = z_{2} . \end{array}$

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

0 0

New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} * \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}$

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations Exemplar Solution

Find $z$ if $? z ? = 4$ and $(z) = \frac{5 π}{6}$ .

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z | = 4 a n d a r g (z) = \frac{5 π}{6} \Rightarrow θ = \frac{5 π}{6} \\ | z | = 4 \Rightarrow r = 4 \\ S o P o l a r f o r m o f z = r [c o s θ + i s i n θ] \\ = 4 [c o s \frac{5 π}{6} + i s i n \frac{5 π}{6}] \\ = 4 [c o s (π - \frac{π}{6}) + i s i n (π - \frac{π}{6})] \\ = 4 [- c o s \frac{π}{6} + i s i n \frac{π}{6}] = 4 [\frac{\sqrt[]{3}}{2} + i . \frac{1}{2}] \\ = - 2 \sqrt[]{3} + 2 i \\ H e n c e, z = - 2 \sqrt[]{3} + 2 i . \end{array}$

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y,$ what is the value of $x^{2} + y^{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + y i \dots (i) \\ T a k i n g c o n j u g a t e o n b o t h s i d e s \\ \frac{{(a^{2} + 1)}^{2}}{2 a + i} = x - y i \dots (i i) \\ M u l t i p l y i n g e q n . (i) a n d (i i) w e h a v e \\ \frac{{(a^{2} + 1)}^{2} {(a^{2} + 1)}^{2}}{(2 a - i) (2 a + i)} = x^{2} + y^{2} \\ \Rightarrow \frac{{(a^{2} + 1)}^{4}}{4 a^{2} - i^{2}} = x^{2} + y^{2} \\ \Rightarrow \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1} = x^{2} + y^{2} \\ H e n c e, t h e v a l u e o f x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1} . \end{array}$

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

0 0

New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

0 0

New answer posted

3 months ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) The order relation is defined on the set of complex numbers.

(ii) Multiplication of a non-zero complex number by $- i$ rotates the point about the origin through a right angle in the anti-clockwise direction.

(iii) For any complex number $z$ , the minimum value of $? z + \frac{1}{z} ?$ is 1.

(iv) The locus represented by $? z - 1 ? = ? z - i ?$ is a line perpendicular to the join of $(1, 0)$ and $(0, 1)$ .

(v) If $z$ is a complex number such that $z \neq 0$ and Re $(z) = 0$ , then Im $(z^{2}) = 0$ .

(vi) The inequality $? z - 4 ? < ? z - 2 ?$ represents the region given by $x > 3$ .

(vii) Let $z_{1}$ and $z_{2}$ be two complex numbers such that $? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$ , then ar

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(i) The order relation is defined on the set of complex numbers.

(ii) Multiplication of a non-zero complex number by $- i$ rotates the point about the origin through a right angle in the anti-clockwise direction.

(iii) For any complex number $z$ , the minimum value of $? z + \frac{1}{z} ?$ is 1.

(iv) The locus represented by $? z - 1 ? = ? z - i ?$ is a line perpendicular to the join of $(1, 0)$ and $(0, 1)$ .

(v) If $z$ is a complex number such that $z \neq 0$ and Re $(z) = 0$ , then Im $(z^{2}) = 0$ .

(vi) The inequality $? z - 4 ? < ? z - 2 ?$ represents the region given by $x > 3$ .

(vii) Let $z_{1}$ and $z_{2}$ be two complex numbers such that $? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$ , then arg $(z_{1} - z_{2}) = 0$ .

(viii) $2$ is not a complex number.

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) C o m p a r i s o n o f t w o p u r e l y i m a g i n a r y c o m p l e x n u m b e r s i s n o t p o s s i b l e . H o w e v e r, t h e t w o \\ p u r e l y i m a g i n a r y r e a l c o m p l e x n u m b e r s c a n b e c o m p a r e d . \\ S o, i t i s' F a l s e' . \\ (i i) L e t z = x + y i \\ z . i = (x + y i) i = x i - y w h i c h r o t a t e s a t a n g l e o f 18 0^{0} \\ S o, i t i s' F a l s e' . \\ (i i i) L e t z = x + y i \\ ∴ | z | + | z + 1 | = \sqrt[]{x^{2} + y^{2}} + \sqrt[]{{(x - 1)}^{2} + y^{2}} \\ T h e v a l u e o f | z | + | z + 1 | i s m i n i m u m w h e n x = 0, y = 0 i . e ., 1 . \\ H e n c e, i t i s' T r u e' . \\ (i v) L e t z = x + y i \\ G i v e n t h a t : | z - 1 | = | z - i | \\ t h e n | x + y i - 1 | = | x + y i - i | \\ \Rightarrow | (x - 1) + y i | = | x - (1 - y) i | \\ \Rightarrow \sqrt[]{{(x - 1)}^{2} + y^{2}} = \sqrt[]{x^{2} + {(1 - y)}^{2}} \\ \Rightarrow {(x - 1)}^{2} + y^{2} = x^{2} + {(1 - y)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} = x^{2} + 1 + y^{2} - 2 y \\ \Rightarrow - 2 x + 2 y = 0 \\ \Rightarrow x - y = 0 w h i c h i s a s t r a i g h t l i n e . \\ S l o p e = 1 \\ N o w e q u a t i o n o f a l i n e t h r o u g h t h e point (1, 0) a n d (0, 1) \\ y - 0 = \frac{1 - 0}{0 - 1} (x - 1) \\ \Rightarrow y = - x + 1 w h o s e s l o p e = - 1 . \\ N o w t h e m u l t i p l i c a t i o n o f t h e s l o p e s o f t w o l i n e s = - 1 * 1 = - 1, \\ S o t h e y a r e p e r p e n d i c u l a r . \\ H e n c e, i t i s' T r u e' . \end{array}$

$\begin{array}{l} (v) L e t z = x + y i, z \neq 0 a n d R e (z) = 0 \\ Since r e a l p a r t i s 0 \Rightarrow x = 0 \\ ∴ z = 0 + y i = y i \\ ∴ I m (z^{2}) = y^{2} i^{2} = - y^{2} w h i c h i s r e a l . \\ H e n c e, i t i s' F a l s e' . \\ (v i) G i v e n t h a t : | z - 4 | < | z - 2 | \\ L e t z = x + y i \\ \Rightarrow | x + y i - 4 | < | x + y i - 2 | \\ \Rightarrow | (x - 4) + y i | < | (x - 2) + y i | \\ \Rightarrow \sqrt[]{{(x - 4)}^{2} + y^{2}} < \sqrt[]{{(x - 2)}^{2} + y^{2}} \\ \Rightarrow {(x - 4)}^{2} + y^{2} < {(x - 2)}^{2} + y^{2} \\ \Rightarrow {(x - 4)}^{2} < {(x - 2)}^{2} \\ \Rightarrow x^{2} - 8 x + 16 < x^{2} - 4 x + 4 \\ \Rightarrow - 8 x + 4 x < - 16 + 4 \\ \Rightarrow - 4 x < - 12 \\ \Rightarrow x > 3 \\ H e n c e, i t i s' T r u e' . \\ (v i i) L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ \Rightarrow | z_{1} + z_{2} | = | z_{1} | + | z00New answer posted 3 months ago Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 12th Chapter Five If a and b are real numbers such that (2 + α)^{4} = a + b α, where α = \frac{- 1 + i \sqrt[]{3}}{2}, then a + b is equal to: 0 Follower 3 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post A alok kumar singh Contributor-Level 10 x^{2} - 3 x + p = 0 α, β, γ, δ in G.P. α + α r = 3 x^{2} - 6 x + q = 0 α r^{2} + α r^{3} = 6 (2) \div (1) \Rightarrow r^{2} = 2 So, \frac{2 q + p}{2 q - p} = \frac{2 r^{5} + r}{2 r^{5} - r} = \frac{2 r^{4} + 1}{2 r^{4} - 1} = \frac{9}{7} 00New answer posted 3 months ago Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 12th Chapter Five Let α, β, be the roots of the equation x^{2} - 4 λ x + 5 = 0 a n d α, γ be the roots of the equation x^{2} - (3 \sqrt[]{2} + 2 \sqrt[]{3}) x + 7 + 3 λ \sqrt[]{3} = 0, If β + γ = 3 \sqrt[]{2}, then {(α + 2 β + γ)}^{2} is equal to. 0 Follower 5 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post P Payal Gupta Contributor-Level 10 α, β are roots of x 2 - 4 λ x + 5 = 0 ∴ α + β = 4 λ a n d α β = 5 Also α, γ are roots of x^{2} - (3 \sqrt[]{2} + 2 \sqrt[]{3}) x + 7 + 3 \sqrt[]{3} λ = 0, λ > 0 ∴ α + γ = 3 \sqrt[]{2} + 2 \sqrt[]{3}, α γ = 7 + 3 \sqrt[]{2 λ} ? α is common root ∴ α^{2} - 4 λ α + 5 = 0 …….(i) And α^{2} - (3 \sqrt[]{2} + 2 \sqrt[]{3}) α + 7 + 3 \sqrt[]{3} λ = 0 ….(ii) From (i) – (ii) ; we get α = \frac{2 + 3 \sqrt[]{3} λ}{3 \sqrt[]{2} + 2 \sqrt[]{3} - 4 λ} ? β + γ = 3 \sqrt[]{2} ∴ {(α + 2 β + γ)}^{2} = {(α + β + β + γ)}^{2} = 98 00❮ 11121314151617181920 ❯Taking an Exam? Selecting a College? Get authentic answers from experts, students and alumni that you won't find anywhere else Sign Up on Shiksha On Shiksha, get access to 65k Colleges 1.2k Exams 687k Reviews 1800k Answers .write-a-review-widget h2.tbSec2 {
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initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedExams","enterPressedHandler":"handleAutoSuggestorEnterPressedExams","mouseClickedHandler":"handleAutoSuggestorEnterPressedExams","backKeyHandler":"handleAutoSuggestorBackKeyPressedExams","objectName":"autoSuggestorInstanceExam","searchBoxId":"searchby-exam","suggestionContainerId":"search-exam-layer","suggestionType":"exam","suggestionCount":8,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","showGrouping":1,"getTrendingSuggestions":1}');
tagsASConfig = '{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","enterPressedHandler":"handleAutoSuggestorEnterPressedTags","mouseClickedHandler":"handleAutoSuggestorEnterPressedTags","objectName":"autoSuggestorInstanceForTags","searchBoxId":"tagSearch","suggestionContainerId":"tagSearch_container","typeAhead":0,"suggestionType":"tag","suggestionCount":10,"autoSuggestorUrl":"\/Tagging\/TaggingDesktop\/getAutoSuggestor","showGrouping":1,"disableCache":1,"removePartialMatch":0}';
initializeAutoSuggestorInstanceSearchV2(tagsASConfig);
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","searchVersion":"2","enterPressedHandler":"handleAutoSuggestorEnterPressedTags","mouseClickedHandler":"handleAutoSuggestorEnterPressedTags","objectName":"autoSuggestorInstanceForTagsQDP","searchBoxId":"tagSearchQDP","suggestionContainerId":"tagSearchQDP_container","typeAhead":0,"suggestionType":"tag","suggestionCount":6,"autoSuggestorUrl":"\/Tagging\/TaggingDesktop\/getAutoSuggestor","showGrouping":1,"disableCache":1,"callBackFunctionAfterBuildingSuggestionContainer":"showSuggestedTags","callBackFunctionOnInputKeysPressed":"showSuggestedTags"}');
}
catch(e) {
console.log(e);
}
}
if(typeof(initializeAutoSuggestorInstancesForCollgeReviews) == 'function') {
initializeAutoSuggestorInstancesForCollgeReviews();
}
if(typeof(initializeAutoSuggestorInstancesForCampusConnect) == 'function') {
initializeAutoSuggestorInstancesForCampusConnect();
}

if(typeof(initializeAutoSuggestorInstanceAlt) == 'function') {
initializeAutoSuggestorInstanceAlt();
}
}
},1000);
}
});
LazyLoadAnADesktopCallback();
}
(function(g,b,d){var c=b.head||b.getElementsByTagName("head"),D="readyState",E="onreadystatechange",F="DOMContentLoaded",G="addEventListener",H=setTimeout;
function f(){ loadJsFilesInParallel(); /*if(typeof CTACallBackSearch == 'function') CTACallBackSearch();*/} H(function(){if("item"in c){if(!c[0]){H(arguments.callee,25);return}c=c[0]}var a=b.createElement("script"),e=false;a.onload=a[E]=function(){if((a[D]&&a[D]!=="complete"&&a[D]!=="loaded")||e){return false}a.onload=a[E]=null;e=true;f()};
a.src="//js.shiksha.ws/public/js/LAB.min.js";c.insertBefore(a,c.firstChild)},0);if(b[D]==null&&b[G]){b[D]="loading";b[G](F,d=function(){b.removeEventListener(F,d,false);b[D]="complete"},false)}})(this,document);lazyLoadCss('//css.shiksha.ws/public/css/build/socialSharingInner.min.5de709.css');var reset_password_token = '';
var reset_usremail = '';
var reset_password = '';
var usergroup = '';

//var registration_context = '';
var user_logged_in_pref_data = "";
var userInfo = null;
var firstTimePageLoad = true;var serviceWorker = "service-worker.js";
if ('serviceWorker' in navigator && typeof serviceWorker == 'string' && serviceWorker.length > 0 && typeof registerServiceWorker == 'function') {
registerServiceWorker(serviceWorker);
}
/* function to bind autosuggestor */
if(typeof(bindAutoSuggestorEvents) == 'function'){
bindAutoSuggestorEvents();
}
/* responsible for showing GNB sub-menu on mouse hover */
if(typeof(activateGnbMouseOver) == 'function'){
activateGnbMouseOver();
}
var homepageMiddlePanel, instituteDetailPageClass, cookieDomain = '.shiksha.com';
$j(document).ready(function(){
shikshaOnreadyEventPC();
getSocialSharingLayer();
});
$j(window).on('load',function(){
shikshaOnloadEventPC();
if(typeof isUserLoggedIn !='undefined' && isUserLoggedIn && typeof getPredictorList != 'undefined' && typeof getPredictorList == 'function'){
getPredictorList();
}
});window.pushNotiConf = {
deviceType : 'desktop'
};
if(document.getElementById("pushNotificationContainer") != null){
var d = new Date();
var dStr = d.getDate()+''+d.getMonth()+''+d.getFullYear()+'v10';
var randomId = Math.floor(Math.random() * 100);

var loadNotificationSDK = function() {
var obj = document.createElement('script');
/* obj.src = 'https://localshiksha.com/public/js/shikshaPushNotification.js?'+dStr+randomId;*/
obj.src = '//js.shiksha.ws/public/js/build/shikshaPushNotification.b0fa7c.js?'+dStr+randomId;
obj.async = true;
document.head.appendChild(obj);
};

var helperObj = document.createElement('script');
/* helperObj.src = 'https://localshiksha.com/pwa/public/js/push-notification-helpers.js?'+dStr;*/
helperObj.src = '//js.shiksha.ws/pwa/public/push-notification-helpers.js?'+dStr+randomId;
helperObj.async = true;
document.head.appendChild(helperObj);

if (typeof helperObj.onload == 'object') {
helperObj.onload = function(e) {
console.log('notification helper sdk loaded');
loadNotificationSDK();
};
} else {
loadNotificationSDK();
}
if(typeof gaTrackEventCustom == 'function'){
window.phpGATrack = gaTrackEventCustom;
}
}setTimeout(function(){if(typeof loadViewsOnPageLoad == 'function'){loadViewsOnPageLoad();} },2000);var multiLocationCourses = [0];if(shikshaProduct=='CMSExam'){
var elem = document.getElementById("cookieAdSlot-d");
elem.parentNode.removeChild(elem);
}facebook Twitter Google +if (document.addEventListener){
document.addEventListener('mouseup', handleShareLayerHide, false);
} else if (document.attachEvent){
document.attachEvent('onmouseup', handleShareLayerHide);
}

function handleShareLayerHide(e)
{
var target = (e && e.target) || (event && event.srcElement);
var obj = $('socialLayer');
if(target!=obj){
obj.style.display='none';
}
}×//Load website tour
$j(document).ready(function(){
window.contentMapping = {"Filters":"Filter rankings by publisher, exams, location, specialization etc.","DEB":"Download details of eligibility, admissions, fees, infra etc.","Shortlist":"Save your shortlist, get updates, college recommendations etc.","Compare":"Compare colleges on ranking, placements, reviews, fees etc.","Reviews":"Get honest reviews from students who studied in this college","Questions":"Get answers from the students currently studying in this college","Institute":"View college details, courses offered & recommendations","Course":"View course details, admission process & insights","Exam":"View Exam Detail","QuestionSearch":"Now you can also search from 5 lakh+ answered questions","DownloadGuide":"Get all details offline, receive important updates & more.","QuestionPapers":"Get prep tips & previous years question papers."};
initializeWebsiteTour('cta','anaDesktopV2',contentMapping);
});
lazyLoadCss('//css.shiksha.ws/public/css/build/quesDiscPosting.min.99ce37.css'); \end{array}$

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