NCERT Solutions Class 11 – Chapter 4 Complex Numbers And Quadratic Equations – Miscellaneous Exercise – GeeksforGeeks

by

NCERT Solutions Class 11 – Chapter 4 Complex Numbers And Quadratic Equations – Miscellaneous Exercise

Last Updated :
19 Apr, 2024

Comments

Improve



Suggest changes

Like Article

Like

<!– –>

<!–


Share

–>


Report

<!– –>


<!–


News
Follow

–>

var video_api_response = JSON.parse(“[]”);

Question 1. Evaluate left[i^{18}+left(frac{1}{i}right)^{25}right]^3

Solution:

displaystyleleft[i^{18}+left(frac{1}{i}right)^{25}right]^3\ =left[i^{4times4+2}+frac{1}{i^{4times6+1}}right]^3\ =left[(i^4)^{4}.i^2+frac{1}{(i^4)^6.i}right]^3\ =left[i^2+frac{1}{i}right]^3 [i^4=1]\ =left[-1+frac{1}{i}timesfrac{i}{i}right]^3 [i^2=-1]\ =left[-1+frac{i}{i^2}right]^3

= [-1 – i]3

= (-1)3 [1 + i]3

= -[13 + i3 + 3 × 1 × i (1 + i)]

= -[1 + i3 + 3i + 3i2]

= -[1 – i + 3i – 3]

= -[2 + 2i]

= 2 – 2i 

Question 2. For any two complex numbers z1 and z2, prove that, Re (z1z2) = Re z1 Re z2 – Im z1 Im z2

Solution:

Let’s assume z1 = x1 + iy1 and z2 = x2 + iy2 as two complex numbers

Product of these complex numbers, z1z2

z1z2 = (x1 + iy1)(x2 + iy2)

= x1(x2 + iy2) + iy1(x2 + iy2)

= x1x2 + ix1y2 + iy1x2 + i2y1y2

= x1x2 + ix1y2 + iy1x2 – y1y2             [i2 = -1]

= (x1x2 – y1y2) + i(x1y2 + y1x2)

Now, 

Re(z1z2) = x1x2 – y1y2

⇒ Re(z1z2) = Rez1Rez2 – Imz1Imz2

Hence, proved.

Question 3. Reduce to the standard form displaystyleleft(frac{1}{1-4i}-frac{2}{1+i}right)left(frac{3-4i}{5+i}right)

Solution:

displaystyleleft(frac{1}{1-4i}-frac{2}{1+i}right)left(frac{3-4i}{5+i}right)=left[frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}right]left[frac{3-4i}{5+i}right]\ = left[frac{1+i-2+8i}{1+i-4i-4i^2}right] left[frac{3-4i}{5+i}right]= left[frac{-1+9i}{5-3i}right] left[frac{3-4i}{5+i}right]\ =  left[frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}right]=frac{33+31i}{28-10i}=frac{33+31i}{2(14-5i)}\ = frac{(33+31i)}{2(14-5i)}timesfrac{14+5i}{14+5i}

On multiplying numerator and denominator by (14+5i)

\ =frac{462+165i+434i+155i^2}{2[(14)^2-(5i)^2]}=frac{307+599i}{2(196-25i^2)}\ =frac{307+599i}{2(221)}=frac{307+599i}{442}=frac{307}{442}+frac{599i}{442}

Hence, this is the required standard form.

Question 4. If x – iy = sqrt{frac{a-ib}{c-id}}  prove that (x2 + y2)2=frac{a^2+b^2}{c^2+d^2}

Solution:

Given:

x – iy = sqrt{frac{a-ib}{c-id}}

=sqrt{frac{a-ib}{c-id}timesfrac{c+id}{c+id}}

On multiplying numerator and denominator by (c+id)

\ =sqrt{frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}

So,

(x – iy)2 = =frac{(ac+bd)+i(ad-bc)}{c^2+d^2}  

x2 – y2 – 2ixy  =frac{(ac+bd)+i(ad-bc)}{c^2+d^2}

On comparing real and imaginary parts, we get

x2 – y2 =  frac{ac+bd}{c^2+d^2} , -2xy = frac{ad-bc}{c^2+d^2}        (1)

(x2 + y2)2 = (x2 – y2)2 + 4x2y2

= left( frac{ac+bd}{c^2+d^2}right)^2+left(frac{ad-bc}{c^2+d^2} right)^2 [Using (1)\ =frac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}\ =frac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}\ = frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^2}\ =frac{(c^2+d^2)(a^2+b^2)}{(c^2+d^2)^2}\ = frac{a^2+b^2}{c^2+d^2}

Hence proved

Question 5. If z1 = 2 – i, z2 = 1 + i, find left|frac{z_1+z_2+1}{z_1-z_2+1}right|

Solution:

Given, z1 = 2 – i, z2 = 1 + i

left|frac{z_1+z_2+1}{z_1-z_2+1}right|=left|frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}right|\ =left|frac{4}{2-2i}right|=left|frac{4}{2(1-i)}right|\ =left|frac{2}{1-i}timesfrac{1+i}{1+i}right|=left|frac{2(1+i)}{1^2-i^2}right|\ =left|frac{2(1+i)}{1+1}right| [i^2=-1]\ =left|frac{2(1+i)}{2}right|\ =|1+i|=sqrt{1^2+1^2}=sqrt2

Hence, the value of left|frac{z_1+z_2+1}{z_1-z_2+1}right| is √2

Question 6. If a + ib =  frac{(x+i)^2}{2x^2+1}, prove that a2 + b2 = frac{(x^2+i)^2}{(2x^2+1)^2}

Solution:

Given:

a + ib = frac{(x+i)^2}{2x^2+1}\ =frac{x^2+i^2+2xi}{2x^2+1}\ =frac{x^2-1+i2x}{2x^2+1}\ =frac{x^2-1}{2x^2+1}+ileft(frac{2x}{2x^2+1}right)

On comparing the real and imaginary parts, we have

a = frac{(x-1)}{2x^2+1}    and b = frac{2x}{2x^2+1}

Therefore,

a2 + b2 = left(frac{x^2-1}{2x^2+1}right)^2+left(frac{2x}{2x^2+1}right)^2\ =frac{x^4+1-2x^2+4x^2}{(2x+1)^2}\ =frac{x^4+1+2x^2}{(2x^2+1)^2}\ =frac{(x^2+1)^2}{(2x^2+1)^2}

Hence, proved,

a2 + b2 = frac{(x^2+1)^2}{(2x^2+1)^2}

Question 7. Let z1 = 2 – i, z2 = -2 + i. Find

(i) Releft(frac{z_1z_2}{overline{z_1}}right)

(ii) Imleft(frac{1}{z_1overline{z_2}}right)

Solution:

(i) Given:

z1 = 2 – i, z2 = -2 + i

(i) z1z2 = (2 – i)(-2 + i) = -4 + 2i + 2i – i2 = -4 + 4i – (-1) = -3 + 4i

overline{z_1} = 2 + i

Therefore,

frac{z_1z_2}{overline{z_1}}=frac{-3+4i}{2+i}

On multiplying numerator and denominator by (2 – i), we get

frac{z_1z_2}{overline{z_1}}=frac{(-3+4i)(2-i)}{(2+i)(2-i)}=frac{-6+3i+8i-4i^2}{2^2+1^2}=frac{-6+11i-4(-1)}{2^2+1^2}\ =frac{-2+11i}{5}=frac{-2}{5}+frac{11}{5}i

On comparing the real parts, we have

Releft(frac{z_1z_2}{overline{z_1}}right)=frac{-2}{5}

(ii) frac{1}{z_1overline{z_2}}=frac{1}{(2-1)(2+i)}=frac{1}{(2)^2+(1)^2}=frac{1}{5}\

On comparing the imaginary part, we get

Imleft(frac{1}{z_1overline{z_2}}right) = 0

Question 8. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Solution:

Let us assume z = (x – iy) (3 + 5i)

z = 3x + xi – 3yi – 5yi2 = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)

Therefore,

overline{z} =(3x + 5y) – i(5x – 3y)

Also given, overline{z}   = -6 – 24i 

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

After equating real and imaginary parts, we get

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

After doing (i) x 3 + (ii) x 5, we have

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of x and y are 3 and –3 respectively.

Question 9. Find the modulus of frac{1+i}{1-i}-frac{1-i}{1+i}

Solution:

frac{1+i}{1-i}-frac{1-i}{1+i}=frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}\ =frac{1+i^2+2i-1-i^2+2i}{1^2+1^2}\ =frac{4i}{2}=2i\ thereforeleft|frac{1+i}{1-i}-frac{1-i}{1+i}right|=|2i|=sqrt{2^2}=2

Question 10. If (x + iy)3 = u + iv, then show that frac{u}{y}+frac{v}{y} = 4(x2 – y2)

Solution:

(x + iy)3 = u + iv

x3 + (iy)3  + 3 × x × iy(x + iy) = u + iv

x3 + i3y3 + 3x2yi + 3xy2 = u + iv

x3 – iy3 + 3x2yi – 3xy2 = u + iv

(x3 – 3xy2) + i(3x2y – y3) = u + iv

On equating real and imaginary parts, we get

u = x3 – 3xy2, v = 3x2y – y3

frac{u}{x}+frac{v}{y}=frac{x^3-3xy^2}{x}+frac{3x^2y-y^3}{y}\ =frac{x(x^2-3y^2)}{x}+frac{y(3x^2y-y)}{y}

= x2 – 3y2 + 3x2 – y2

= 4x2 – 4y2

= 4(x2 – y2)

thereforefrac{u}{x}+frac{v}{y}=4(x^2-y^2)

Hence proved

Question 11. If α and β are different complex numbers with |β| = 1, then find left|frac{beta-alpha}{1-overline{alpha}beta}right|

Solution:

Assume α = a + ib and β = x + iy

Given: |β| = 1

So, sqrt{x^2+y^2}=1

= x2 + y2 = 1            ….(1)

left|frac{beta-alpha}{1-overline{alpha}beta}right|=left|frac{(x+iy)(a+ib)}{1-(a-ib)(x+iy)}right|\ =left|frac{(x-a)+i(y-b)}{1-(ax+aiy-ibx+by)}right|\ =left|frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}right|\ =left|frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}right| left[left|frac{z_1}{z_2}right|=left|frac{z_1}{z_2}right|right]\ =frac{sqrt{(x-a)^2+(y-b)^2}}{sqrt{(1-ax-by)^2+(bx-ay)^2}}\ =frac{sqrt{x^2+a^2-2ax+y^2+b^2-2by}}{sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-2by+b^2x^2+a^2y^2-2abxy}}\ =frac{sqrt{(x^2+y^2)+a^2+b^2-2ax-2by}}{sqrt{1+a^2(x^2+y^2)+b^2(y^2+x^2)-2ax-2by}}\ =frac{sqrt{1+a^2+b^2-2ax-2by}}{sqrt{1+a^2+b^2-2ax-2by}} [Using (1)]  

= 1

thereforeleft|frac{beta-alpha}{1-overline{alpha}beta}right|=1

Question 12. Find the number of non-zero integral solutions of the equation |1 – i|x = 2x

Solution:

|1 – i|x = 2t

(sqrt{1^2+(-1)^2})^x=2^x\ (sqrt2)^x=2^x\ 2^{frac{x}{2}}=2^x\ frac{x}{2}=x

x = 2x

2x – x = 0

Thus, ‘0’ is the only integral solution of the given equation.

Therefore, the number of non-zero integral solutions of the given equation is 0.

Question 13. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

Solution:

Given:

(a + ib)(c + id)(e + if)(g + ih) = A + iB

Therefore,

|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|

= |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|

sqrt{a^2+b^2}timessqrt{c^2+d^2}timessqrt{e^2+f^2}timessqrt{g^2+h^2}=sqrt{A^2+B^2}

On squaring both sides, we get

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2

Hence, proved.

Question 14. If, then find the least positive integral value of m. left(frac{1+i}{1-i}right)^m=1

Solution:

left(frac{1+i}{1-i}right)^m=1

left(frac{1+i}{1-i}timesfrac{1+i}{1+i}right)^m=1\ left(frac{(1+i)^2}{1^2+1^2}right)^m=1\ left(frac{1-1+2i}{2}right)^m=1\ left(frac{2i}{2}right)^m=1

im = 1

Hence, m = 4k, where k is some integer.

Hence, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).

const rawVideoDataMap = null;

// Convert it into a JS Map
const videoDataMap = new Map(Object.entries(rawVideoDataMap));

document.addEventListener(‘DOMContentLoaded’, function () {
let currentlySelectedTile = document.getElementById(“video-info-container-title”);
let firstVideoH2;
let firstSelectedVideo
var videoPlayer = document.getElementById(‘video-iframe-container’);
if (!videoPlayer) return;

var videoIframe = videoPlayer.querySelector(‘iframe’);
if (!videoIframe) return;

// Safely inject PHP data as JSON into JavaScript
var resp = null;

function postToIframeWhenReady(iframe, data, maxAttempts = 20, intervalTime = 100) {
let attempts = 0;
const interval = setInterval(() => {
if (iframe && iframe.contentWindow) {
try {
iframe.contentWindow.postMessage(data, ‘*’);
clearInterval(interval);
} catch (err) {
console.warn(“❌ Failed to postMessage, retrying…”);
}
}
if (++attempts >= maxAttempts) {
console.warn(“⚠️ postMessage failed: iframe not ready after multiple attempts.”);
clearInterval(interval);
}
}, intervalTime);
}

// Call the function right after iframe is confirmed to exist
videoIframe.onload = function () {
postToIframeWhenReady(videoIframe, { resp: resp, inView: false });
};

const playlistContainer = document.getElementById(“video-playlist-container”);
if(playlistContainer) {
firstSelectedVideo = playlistContainer.querySelectorAll(‘.playlist-video-tile’)[0];
if(firstSelectedVideo) {
firstVideoH2 = firstSelectedVideo.querySelector(‘h2’);
firstVideoH2.style.setProperty(‘color’, ‘var(–discussion-button-color)’, ‘important’);
}
}

function changeCurrentPlayingVideoTitle(tile, autoplayed = true) {
const playlistVideoTileHeading = tile.querySelector(‘h2’);
if (playlistVideoTileHeading) {
if(!autoplayed && firstVideoH2.innerText !== playlistVideoTileHeading.innerText) {
firstVideoH2.style.setProperty(‘color’, ‘#EC4E20’, ‘important’);
}
if(autoplayed) {
let siblingElement = tile.previousElementSibling;
const siblingElementH2 = siblingElement.querySelector(‘h2’);
siblingElementH2.style.setProperty(‘color’, ‘#EC4E20’, ‘important’);
firstVideoH2 = playlistVideoTileHeading;
}
playlistVideoTileHeading.style.setProperty(‘color’, ‘var(–discussion-button-color)’, ‘important’);
if (currentlySelectedTile) {
currentlySelectedTile.innerText = playlistVideoTileHeading.innerText;
}
}
}
// Inject the video data map as a JS object
window.addEventListener(‘message’, function (event) {
if (event.data.type === ‘AUTOPLAYED_NEXT_VIDEO’ && event.data.autoplayPlaylistEnabled) {

// Changing the current video index on autoplaying the next video from playlist
const allVideoTiles = Array.from(document.querySelectorAll(‘.playlist-video-tile’));
changeCurrentPlayingVideoTitle(allVideoTiles[event.data.currentVideoIndex]);
}
});

if(playlistContainer) {
playlistContainer.addEventListener(‘click’, (event) => {
const tile = event.target.closest(‘.playlist-video-tile’);

// Find the index of the clicked tile
const allTiles = Array.from(playlistContainer.querySelectorAll(‘.playlist-video-tile’));
const currentIndex = allTiles.findIndex((el) => el === tile);

if (tile) {
const videoId = tile.getAttribute(‘data-video-id’);
const videoData = videoDataMap.get(videoId);
videoIframe.contentWindow.postMessage({ playlist: [videoData], inView: true, clickedVideoFromPlaylist: true, currentIndex: currentIndex }, ‘*’);

// handleVideoTitle([videoData]);
changeCurrentPlayingVideoTitle(tile, false);
firstVideoH2 = tile.querySelector(‘h2’);
}
});
}
});

var video_api_response = JSON.parse(“[]”);

S

shristiag18

Improve

S

shristiag18

<!–


News

–>

<!– Like Article –>
Improve

Article Tags :

Explore

NCERT Exemplar Solutions (Q11 to 20) Chemical Bonding and Molecular Structure ➡️

📚 Buy Study Material & Join Our Coaching

For premium study materials specially designed for NDA Exam, visit our official study material portal:
👉 https://publishers.anandclasses.co.in/

For JEE/NEET Notes : Visit https://anandclasses.in/

To enroll in our offline or online coaching programs, visit our coaching center website:
👉 https://anandclasses.co.in/

📞 Call us directly at: +91-94631-38669

💬 WhatsApp Us Instantly

Need quick assistance or want to inquire about classes and materials?

📲 Click below to chat instantly on WhatsApp:
👉 Chat on WhatsApp

🎥 Watch Video Lectures

Get access to high-quality video lessons, concept explainers, and revision tips by subscribing to our official YouTube channel:
👉 Neeraj Anand Classes – YouTube Channel

RELATED TOPICS

Download free NCERT Solutions PDF for Class 11 Maths Chapter 4: Complex Numbers & Quadratic Equations

NCERT Solutions Complex Numbers & Quadratic Equations Miscellaneous Exercise

NCERT Solutions Class 11 – Chapter 4 Complex Numbers And Quadratic Equations – Miscellaneous Exercise – GeeksforGeeks

Download free NCERT Solutions PDF for Class 11 Maths Chapter 4: Complex Numbers & Quadratic Equations

NCERT Solutions Complex Numbers & Quadratic Equations Ex-4.1 free pdf download

NCERT Exemplar Solutions (Q11 to 20) Chemical Bonding and Molecular Structure

NCERT Solutions for Class 11 Chemistry Chapter – Chemical Bonding and Molecular Structure

NCERT Exemplar Solutions (Q1 to 10) Chemical Bonding and Molecular Structure

NCERT Solutions for Class 11 Chemistry Chapter – Chemical Bonding and Molecular Structure

NCERT (Q4.31 to Q4.40) Chemical Bonding and Molecular Structure Solutions pdf download free Notes

NCERT Solutions for Class 11 Chemistry Chapter – Chemical Bonding and Molecular Structure

NCERT (Q4.11 to Q4.20) Chemistry Solutions Chemical Bonding pdf download free

NCERT Solutions for Class 11 Chemistry Chapter – Chemical Bonding and Molecular Structure

Chemical Bonding NCERT (Q4.20 to Q4.30) Solutions pdf download free Notes

NCERT Solutions for Class 11 Chemistry Chapter – Chemical Bonding and Molecular Structure

NCERT (Q4.1 to Q4.10) Chemistry Solutions Chemical Bonding and Molecular Structure

Chemical Bonding And Molecular Structure Class 11 JEE NEET Study Material pdf download free

Why Ice (less dense) floats over Water and Density of Water is maximum at 4°C

📞 Call Us Now

Speak directly with our academic experts to know more about admissions, study material, or batch schedules.

Call: +91-94631-38669

💬 Chat on WhatsApp

Get instant responses to your questions about coaching programs and study resources.

WhatsApp Now

▶️ Watch Video Lectures

Explore high-quality video lessons and concept explanations by Neeraj Anand Sir.

Visit YouTube Channel
© 2025 Anand Classes • Published by Anand Technical Publishers
Privacy Policy Terms Contact

“Your Trusted Partner for JEE, NEET & NDA Preparation”

"Anand Technical Publishers, under Anand Classes, is dedicated to publishing high-quality, exam-oriented books for JEE, NEET, NDA and Board exams—crafted by expert educators to ensure concept clarity and competitive success.