Complex Numbers and Quadratic Equations: Class 11 Maths Chapter 4 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

60 min
1 simulation
27 practice questions
19 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki x² + 1 = 0 solve karne ke liye real numbers ko extend karna kyun zaroori hai, aur complex number a + ib ko uske real aur imaginary parts se describe karo.
Complex numbers ki equality ka use karke, real aur imaginary parts compare karke unknown real quantities nikalo.
Complex numbers ko add, subtract, multiply aur divide karo, aur batao ki yeh operations kaun se laws follow karte hain.
i ki kisi bhi integer power ko simplify karo aur negative real numbers ke square roots ko i ke multiples ki tarah likho, √a√b = √(ab) ka galat use kiye bina.
Complex number ka modulus, conjugate aur multiplicative inverse nikalo, aur unki properties se expressions simplify karo aur results prove karo.
Complex numbers ko Argand plane mein represent karo: modulus ko origin se distance ki tarah aur conjugate ko real axis mein reflection ki tarah padho.

NCERT Mathematics Class 11, Chapter 4, pp. 76–88 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

4.1 Introduction

Phone ke calculator mein $\sqrt{- 9}$ type karo, to zyaadatar woh Error dikhayega. Real numbers ke andar calculator sahi hai: kisi bhi real number ka square zero ya positive hota hai, isliye aisa koi real number nahi jiska square $- 9$ ho. Is chapter mein hum ek bada number system banayenge jismein aise square roots exist karte hain, aur usmein utni hi aasaani se calculate karna seekhenge jitna ordinary numbers ke saath karte hain.

Yeh pehli baar nahi hai ki number system bada hua hai. Jab bhi kisi simple equation ka koi answer nahi mila, mathematicians ne haar maanne ke bajaye system ko bada kar diya. Fig. 4.1 is safar ko dobara dikhata hai: $x + 5 = 2$ ke liye negative integers chahiye, $3 x = 2$ ke liye fractions aur $x^{2} = 2$ ke liye irrational numbers. Equation $x^{2} + 1 = 0$ is seedhi ka agla step hai.

5/5 $x^{2} = - 1$ ka koi real answer nahi hai. Complex numbers se $x = \pm i$ milta hai.

Fig. 4.1Har naye tarah ka number ek aisi equation ka answer deta hai jo purane numbers nahi de paate the.

Paanch nested rounded boxes, sabse andar natural numbers aur sabse bahar complex numbers. Step 1 mein sirf natural numbers 1, 2, 3 waghairah dikhte hain. Step 2 unke chaaron taraf integers jodta hai, right mein equation x + 5 = 2 aur uske answer x = −3 ke saath. Step 3 rational numbers jodta hai, 3x = 2 aur x = 2/3 ke saath. Step 4 real numbers jodta hai, x squared = 2 aur x = ± root 2 ke saath. Step 5 complex numbers ko sabse bahar wale box ki tarah jodta hai, x squared = −1 aur x = ± i ke saath.

Ek practical wajah bhi hai. Class 10 mein tumne $a x^{2} + b x + c = 0$ ko quadratic formula se solve kiya tha aur jab bhi $b^{2} - 4 a c < 0$ hota, tumhe rukna padta tha. Jab negative numbers ke bhi square roots hon, to formula har baar kaam karta hai. Electrical engineers power grid mein alternating current ko describe karne ke liye roz inhi numbers ka use karte hain, aur signal engineers phone calls le jaane wali radio waves ko process karne ke liye inka use karte hain.

4.2 Complex Numbers

Hum seedha ek bold step lete hain. Hum ek naya number $i$ introduce karte hain, jise iota padhte hain aur jiska square $- 1$ hai, aur tay karte hain ki algebra ka har woh rule rakhenge jo ab bhi sense banata hai. Yeh $i$ imaginary unit hai.

Imaginary unit

i = \sqrt{- 1}, i^{2} = - 1

(4.1)

Imaginary naam historical hai: yeh numbers bhi baaki numbers jitne hi achhi tarah defined hain. $i$ ko real numbers ke saath milane se ek poori nayi family milti hai, complex numbers (सम्मिश्र संख्याएँ).

Definition

Complex number

Ek number $z = a + i b$ jo do real numbers $a$ aur $b$ aur imaginary unit $i$ se bana hai. $a$ ko hum $z$ ka real part ( $Re z$ ) kehte hain aur $b$ ko uska imaginary part ( $Im z$ ).

Example $z = - 4 + \frac{3}{2} i$ ke liye: $Re z = - 4$ aur $Im z = \frac{3}{2}$ .

Dhyaan do ki dono parts real numbers hain. $3 - 2 i$ ka imaginary part $- 2$ hai, $- 2 i$ nahi, jabki uska real part $3$ hai. Fig. 4.2 number ko $3 + (- 2) i$ likh kar yeh saaf dikhata hai. Tum $a + i b$ ya $a + b i$ likh sakte ho; surd ho to $\sqrt{3} i$ ke bajaye $i \sqrt{3}$ likhna behtar hai, taaki koi use $\sqrt{3 i}$ na padh le.

2/3 $Re z = 3$ aur $Im z = - 2$ ; $i$ kisi mein bhi shaamil nahi.

Fig. 4.2

z = 3 - 2 i

asal mein

3 + (- 2) i

hai: dono parts real numbers hain.NCERT 4.2 · p. 76

Number z = 3 + (−2)i bade akshar mein likha hai. Step 2 mein 3 box mein hai aur uska label real part, Re z = 3 hai, aur brackets wala −2 dashed outline ke box mein hai jiska label imaginary part, Im z = −2 hai; letter i dono boxes ke bahar rehta hai. Step 3 mein neeche do chips aate hain: 5 + 0i, label real, aur 0 + 4i, label purely imaginary.

Do tarah ke complex numbers ko alag naam milta hai. Jab $b = 0$ ho, to number $a + 0 i$ bas real number $a$ hai, isliye har real number ek complex number bhi hai. Jab $a = 0$ aur $b \neq 0$ ho, to number $0 + i b = i b$ ko purely imaginary kehte hain, jaise $7 i$ ya $- \frac{1}{2} i$ . Saare complex numbers ke set ko $𝐂$ likhte hain, aur $𝐑 \subset 𝐂$ .

Do complex numbers kab same hote hain? Theek tab, jab woh part by part match karein. Yeh rule complex numbers ke beech ki ek equation ko real numbers ke beech ki do equations mein badal deta hai, aur yeh trick hum baar baar use karenge.Yeh kehne ka koi sensible tareeka nahi hai ki ek complex number doosre se bada hai. $3 + i > 2$ jaise statements ka koi matlab nahi; sirf moduli compare kiye ja sakte hain, jo real hote hain.

Equality

\begin{array}{c} a + i b = c + i d \\ ⟺ a = c aur b = d \end{array}

(4.2)

jahan

$a, b, c, d$: real numbers

Example 4.1NCERT Example 1 · p. 77

Parts ko compare karna

Woh real numbers $x$ aur $y$ nikalo jinke liye $(3 x + 2 y) + i (x - y) = 12 - i .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Im (5 - 7 i)

kya hai?

- 7 i

7

- 7

5

Answer dekho

Answer (C) $- 7$

Imaginary part woh real number hai jo

i

se multiply hota hai, sign ke saath:

- 7

- 7 i

likhna sabse common galti hai;

Im z

mein kabhi

i

nahi hota.

4.3 Complex Numbers ka Algebra

Is poore section ka working rule chhota sa hai: $i$ ke saath bilkul waise hi calculate karo jaise $x$ jaise kisi letter ke saath karte ho, aur jab bhi $i^{2}$ aaye, use $- 1$ se replace kar do. Tab neeche ka har operation phir se $a + i b$ form ka number deta hai, aur arithmetic ke jaane-pehchaane laws bane rehte hain.

4.3.1 Do complex numbers ka addition

Do complex numbers add karne ke liye real parts ko add karo aur imaginary parts ko add karo, bilkul waise hi jaise tum yahan like terms collect karte ho: $(4 - 3 x) + (- 1 + 7 x) .$

Sum

\begin{array}{c} (a + i b) + (c + i d) \\ = (a + c) + i (b + d) \end{array}

(4.3)

Jaise, $(4 - 3 i) + (- 1 + 7 i) = 3 + 4 i .$ Kyunki har part ek ordinary real number ki tarah add hota hai, addition woh saare laws rakhta hai jo tum jaante ho: yeh closed hai (sum phir se complex hota hai), commutative aur associative hai. Number $0 = 0 + 0 i$ additive identity hai, kyunki $z + 0 = z$ , aur har $z = a + i b$ ka ek additive inverse $- z = - a - i b$ hota hai, jismein $z + (- z) = 0$ . Table 4.1 in laws ko multiplication ke laws ke saath list karta hai.

4.3.2 Do complex numbers ka difference

Subtraction additive inverse ke through define hota hai: $z_{1} - z_{2} = z_{1} + (- z_{2}) .$ Practice mein tum real parts ko subtract karte ho aur imaginary parts ko subtract karte ho.

Kya subtraction commutative hai? Ise $z_{1} = 5 + 2 i$ aur $z_{2} = 3 - 4 i$ se test karo. $- z_{2} = - 3 + 4 i$ add karne se milta hai $z_{1} - z_{2} = 2 + 6 i,$ lekin doosre order mein $z_{2} - z_{1} = - 2 - 6 i .$ Dono answers ek doosre ke negative hain, isliye subtraction commutative nahi hai, bilkul real numbers ki tarah.

4.3.3 Do complex numbers ka multiplication

Do complex numbers ko waise hi multiply karo jaise do binomials ko karte ho, phir $i^{2} = - 1$ se simplify karo. $(a + i b) (c + i d)$ ko expand karne par chaar terms aate hain, Fig. 4.3 ke har cell ke liye ek.

Product

\begin{aligned} (a + i b) (c + i d) \\ = (a c - b d) \\ + i (a d + b c) \end{aligned}

(4.4)

4/4Real cells se $a c - b d$ milta hai; imaginary cells se $a d + b c$ .

Fig. 4.3Chaar products, aur jis product mein

i^{2}

hai uska sign badal jaata hai.NCERT 4.3.3 · p. 78

Ek grid jiske upar c aur + id hain aur side mein a aur + ib. Step 1 khaali grid dikhata hai. Step 2 cells mein ac, iad, ibc aur i squared bd bharta hai. Step 3 mein i squared = −1 label wala arrow aakhri cell ki taraf point karta hai, jo ab −bd padhta hai. Step 4 cells ac aur −bd ko real ki tarah shade karta hai aur cells iad aur ibc ko, dashed borders ke saath, imaginary ki tarah, aur neeche result (ac − bd) + i(ad + bc) likhta hai.

Numbers ke saath: $(4 + 3 i) (2 - 5 i) = 8 - 20 i + 6 i - 15 i^{2} .$ Aakhri term $- 15 i^{2} = + 15$ hai, isliye product $23 - 14 i$ hai. (4.4) ko yaad karne ki zaroorat nahi: expand karo, phir $i^{2}$ ko replace karo.

Table 4.1Complex numbers u, v, w ke liye addition aur multiplication ke laws

Law	Addition ke liye	Multiplication ke liye
Closure	$u + v$ complex hai	$u v$ complex hai
Commutative	$u + v = v + u$	$u v = v u$
Associative	$(u + v) + w = u + (v + w)$	$(u v) w = u (v w)$
Identity	$0 = 0 + 0 i$ , $u + 0 = u$	$1 = 1 + 0 i$ , $u \cdot 1 = u$
Inverse	$- u$ , jismein $u + (- u) = 0$	$u \neq 0$ ke liye $u^{- 1}$ , jismein $u u^{- 1} = 1$

Dono operations distributive law se jude hain: $z_{1} (z_{2} + z_{3}) = z_{1} z_{2} + z_{1} z_{3}$ aur $(z_{1} + z_{2}) z_{3} = z_{1} z_{3} + z_{2} z_{3} .$ Har law ko $z_{1}, z_{2}, z_{3}$ ko $a + i b$ ki form mein likh kar aur parts compare karke check kiya ja sakta hai; ab se hum inhe freely use karenge. Sirf multiplicative inverse mein asli mehnat lagti hai.

DerivationReciprocal nikalna

$0$ ko chhod kar har complex number ka ek multiplicative inverse hota hai. $z = a + i b \neq 0$ ke liye hum aise real $x, y$ dhoondhte hain jinke liye $(a + i b) (x + i y) = 1 + 0 i .$ (4.4) se expand karke parts compare karne par do linear equations milti hain.

\begin{aligned} a x - b y & = 1 \\ b x + a y & = 0 \end{aligned}

Pehli equation ko $a$ se aur doosri ko $b$ se multiply karke add karo: $(a^{2} + b^{2}) x = a$ . Pehli ko $- b$ se aur doosri ko $a$ se multiply karke add karo: $(a^{2} + b^{2}) y = - b$ . Kyunki $z \neq 0$ , numbers $a$ aur $b$ dono zero nahi hain, isliye $a^{2} + b^{2} > 0$ aur hum divide kar sakte hain.

Multiplicative inverse

\begin{aligned} z^{- 1} & = \frac{a}{a^{2} + b^{2}} \\ + i \frac{- b}{a^{2} + b^{2}} \end{aligned}

(4.5)

Ise $\frac{1}{z}$ bhi likhte hain. $z \cdot z^{- 1}$ ko multiply karke dekhne se confirm hota hai ki answer $1$ hai.

4.3.4 Do complex numbers ka division

Division ka matlab hai inverse se multiply karna. $z_{2} \neq 0$ ke liye hum define karte hain $\frac{z_{1}}{z_{2}} = z_{1} \cdot \frac{1}{z_{2}},$ aur (4.5) se $\frac{1}{z_{2}}$ milta hai.

Example

Ek complex number ko doosre se divide karna

$10 + 5 i$ ko $1 + 2 i$ se divide karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Section 4.4 ek tez raasta deta hai jismein koi formula nahi chahiye: upar aur neeche dono ko neeche wale number ke conjugate se multiply karo.

4.3.5 i ki Power

$i$ se baar baar multiply karne par sirf chaar values milti hain. $i^{2} = - 1$ se humein $i^{3} = i^{2} \cdot i = - i$ aur $i^{4} = (i^{2})^{2} = 1$ milte hain. Phir $i^{5} = i^{4} \cdot i = i$ , aur cycle phir se shuru ho jaata hai (Fig. 4.4).

Negative powers is fact se nikalti hain: $i \cdot (- i) = - i^{2} = 1,$ yaani $i^{- 1} = - i$ . Isliye $i^{- 2} = (- i)^{2} = - 1,$ $i^{- 3} = i^{- 2} \cdot i^{- 1} = i$ aur $i^{- 4} = 1$ . Peeche ki taraf jaane par wahi chaar values ulte order mein aati hain.

i ki powers

\begin{aligned} i^{4 k} & = 1 \\ i^{4 k + 1} & = i \\ i^{4 k + 2} & = - 1 \\ i^{4 k + 3} & = - i \end{aligned}

(4.6)

jahan

$k$: koi bhi integer, positive, negative ya zero

3/4 $i^{3} = - i$ aur $i^{4} = 1$ : exponent ko $4$ se divide karne par jo remainder bachta hai, wahi station chunta hai.

Fig. 4.4

i

se har multiplication ek station anticlockwise aage le jaata hai; chaar moves mein tum wapas ghar pahunch jaate ho.NCERT 4.3.5 · p. 79

Ek ring par chaar stations hain: right mein 1, upar i, left mein −1 aur neeche −i. Step 1 mein 1 se i tak times i label wala arrow banta hai, aur centre mein i to the power 1 = i. Step 2 i se −1 tak arrow jodta hai, i squared = −1 ke saath. Step 3 loop ko −i se hote hue wapas 1 tak poora karta hai, i to the power 4 = 1 dikhata hai, aur stations par exponents 4k, 4k + 1, 4k + 2 aur 4k + 3 ke labels lagata hai. Step 4 loop ko fade karta hai aur 1 se −i tak peeche jaata hua divide by i label wala dashed arrow banata hai, i to the power −1 = −i ke saath.

To $i^{n}$ simplify karne ke liye $n$ ko $4$ se divide karo aur remainder $r \in {0, 1, 2, 3}$ rakho; phir $i^{n} = i^{r}$ . Yeh rule negative $n$ ke liye bhi kaam karta hai, bas remainder $0$ aur $3$ ke beech lena hai, jaise $- 22 = 4 (- 6) + 2$ .

Example

Badi powers ko kaabu karna

Simplify karo (a) $i^{101}$ , (b) $i^{- 22}$ aur (c) $i + i^{2} + i^{3} + i^{4}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.3.6 Negative real number ke square roots

Kyunki $i^{2} = - 1$ aur saath hi $(- i)^{2} = (- 1)^{2} i^{2} = - 1,$ number $- 1$ ke do square roots hain, $i$ aur $- i$ . Symbol $\sqrt{- 1}$ inmein se ek, $i$ , ke liye reserved hai, bilkul waise jaise $\sqrt{9}$ ka matlab $3$ hai, $- 3$ nahi.

Yahi baat har negative number ke liye sach hai. Agar $a > 0$ , to $(\pm i \sqrt{a})^{2} = a i^{2} = - a,$ isliye $- a$ ke do square roots $\pm i \sqrt{a}$ hain, aur symbol $\sqrt{- a}$ hamesha $i \sqrt{a}$ ko denote karta hai. Isliye $\sqrt{- 49} = 7 i$ aur $\sqrt{- 5} = i \sqrt{5}$ .

Ab ek trap jismein bahut se students phans jaate hain (Fig. 4.5). Jaana-pehchaana rule $\sqrt{a} \sqrt{b} = \sqrt{a b}$ tab fail ho jaata hai jab $a$ aur $b$ dono negative hon. Baaki har case mein yeh safe hai: dono numbers non-negative hon, ya ek positive aur ek negative ho (aur agar koi bhi $0$ hai, to dono sides $0$ hain).

3/3 $\sqrt{(- 4) (- 9)} = \sqrt{36} = 6$ likhne se galat sign milta hai.

Fig. 4.5Multiply karne se pehle

\sqrt{- a}

i \sqrt{a}

mein badlo.NCERT 4.3.6 · p. 80

Upar product root of −4 times root of −9 hai. Step 1 sirf question dikhata hai. Step 2 left mein ek solid green box kholta hai jiska heading 'pehle roots' hai: 2i times 3i, phir 6 i squared, phir −6, tick aur 'sahi' word ke saath. Step 3 right mein ek dashed box kholta hai jiska heading 'pehle merge' hai: root of (−4)(−9), phir root 36, phir 6, jo kata hua hai aur 'galat' mark kiya gaya hai.

Koi chaalaak jugaad is rule ko nahi bacha sakta. Agar yeh $a = b = - 1$ ke liye sahi hota, to humein milta $i^{2} = \sqrt{- 1} \sqrt{- 1} = \sqrt{(- 1) (- 1)} = \sqrt{1} = 1,$ jo $i^{2} = - 1$ ke khilaaf hai. Safe aadat yeh hai ki kuch bhi aur karne se pehle har $\sqrt{- a}$ ko $i \sqrt{a}$ ki tarah dobara likho.

4.3.7 Identities

Real numbers ke liye jo algebraic identities tum jaante ho, woh sirf commutative, associative aur distributive laws se prove hui thi. Complex numbers bhi yahi laws follow karte hain, isliye yahi identities saare complex $z_{1}$ aur $z_{2}$ ke liye sahi hain. Yahan ek proof hai jismein har law ka naam diya gaya hai.

TheoremSquares ka difference

Saare complex numbers $z_{1}$ aur $z_{2}$ ke liye, $z_{1}^{2} - z_{2}^{2} = (z_{1} + z_{2}) (z_{1} - z_{2}) .$

Squares aur cubes

\begin{aligned} (z_{1} \pm z_{2})^{2} \\ = z_{1}^{2} \pm 2 z_{1} z_{2} + z_{2}^{2} \\ (z_{1} \pm z_{2})^{3} \\ = z_{1}^{3} \pm 3 z_{1}^{2} z_{2} \\ + 3 z_{1} z_{2}^{2} \pm z_{2}^{3} \end{aligned}

(4.7)

(4.7) se cube expand karna teen brackets multiply karne se zyada tez aur safe hai, jaisa Example 4.3 dikhata hai. Usse pehle do warm-up products.

Example 4.2NCERT Example 2 · p. 80

Imaginary numbers ke products

$a + i b$ ki form mein likho: (i) $(6 i) (- \frac{2}{3} i)$ (ii) $(- 2 i)^{3} {(\frac{1}{4} i)}^{2}$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.3NCERT Example 3 · p. 81

Binomial ka cube

$(2 + 3 i)^{3}$ ko $a + i b$ ki form mein express karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.4NCERT Example 4 · p. 81

Product mein negatives ke square roots

$(\sqrt{5} + \sqrt{- 3}) (\sqrt{5} - 2 \sqrt{- 3})$ ko $a + i b$ ki form mein express karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.4 Complex Number ka Modulus aur Conjugate

Har complex number se jude do numbers division ko aasaan banate hain aur Section 4.5 mein complex numbers ko ek geometric matlab dete hain.

Definition

Modulus

$z = a + i b$ ka modulus, jise $∣ z ∣$ likhte hain, $\sqrt{a^{2} + b^{2}}$ hai: ek real number jo kabhi negative nahi hota.

Example $∣ 5 - 12 i ∣ = \sqrt{25 + 144} = 13$

Definition

Conjugate

$z = a + i b$ ka conjugate $\overset{ˉ}{z} = a - i b$ hai: real part waisa hi rakho aur imaginary part ka sign ulta kar do.

Example $\overline{4 + 7 i} = 4 - 7 i$

Modulus aur conjugate

\begin{array}{c} ∣ z ∣ = \sqrt{a^{2} + b^{2}} \\ \overset{ˉ}{z} = a - i b \end{array}

(4.8)

Kuch aur examples: $∣ - 2 + i ∣ = \sqrt{5}$ , $\overline{- 3 i} = 3 i$ aur $\overline{6} = 6$ , kyunki real number khud apna conjugate hota hai. Real number $a$ ke liye $∣ a + 0 i ∣ = \sqrt{a^{2}} = ∣ a ∣,$ isliye modulus (मापांक) usi absolute value ko aage badhata hai jo tum pehle se jaante ho. Conjugate do baar lene par original number wapas mil jaata hai: $\overline{\overline{z}} = z$ .

Ek number aur uska conjugate ek kaam ki jodi banate hain (Fig. 4.6). Unka sum $2 a$ real hai, unka difference $2 i b$ purely imaginary hai, aur unka product real hai aur kabhi negative nahi hota.

4/4 $z \overset{ˉ}{z} = a^{2} - (i b)^{2} = a^{2} + b^{2},$ jo $∣ z ∣^{2}$ hai.

Fig. 4.6Ek number aur uske conjugate ko add, subtract aur multiply karna.NCERT 4.4 · p. 81

Left mein ek card par z = a + ib ke neeche z-bar = a − ib hai. Step 1 sirf jodi dikhata hai. Step 2 terms + ib aur − ib ko kaat deta hai, aur right mein ek row z + z-bar = 2a padhti hai, label real. Step 3 iske bajaye dono a terms ko kaatta hai, aur doosri row z − z-bar = 2ib padhti hai, label purely imaginary. Step 4 kaatne ke nishaan hata deta hai aur teesri row jodta hai, z times z-bar = a squared + b squared, label 'real, negative nahi'.

Number aur uske conjugate ka product

\begin{array}{c} z \overset{ˉ}{z} = (a + i b) (a - i b) \\ = a^{2} + b^{2} = ∣ z ∣^{2} \end{array}

(4.9)

(4.9) ko reciprocal (4.5) se compare karo: uska numerator $a - i b = \overset{ˉ}{z}$ hai aur denominator $a^{2} + b^{2} = ∣ z ∣^{2}$ . Isse inverse ko ek saaf-suthri form milti hai. Isse division ka bhi ek aasaan method milta hai: numerator aur denominator dono ko denominator ke conjugate se multiply karo, jisse denominator real number $∣ z_{2} ∣^{2}$ ban jaata hai.

Inverse aur quotient

\begin{array}{c} z^{- 1} = \frac{\overset{ˉ}{z}}{∣ z ∣^{2}} \\ \frac{z_{1}}{z_{2}} = \frac{z_{1} {\overset{ˉ}{z}}_{2}}{∣ z_{2} ∣^{2}} \end{array}

(4.10)

jahan

$z, z_{2}$: zero se alag complex numbers

PropertyModulus aur conjugate ke rules

Complex numbers $z_{1}$ aur $z_{2}$ ke liye: (i) $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣ ∣ z_{2} ∣;$ (ii) agar $z_{2} \neq 0$ to $∣ \frac{z_{1}}{z_{2}} ∣ = \frac{∣ z_{1} ∣}{∣ z_{2} ∣}$ ; (iii) $\overline{z_{1} z_{2}} = {\overset{ˉ}{z}}_{1} {\overset{ˉ}{z}}_{2};$ (iv) $\overline{z_{1} \pm z_{2}} = {\overset{ˉ}{z}}_{1} \pm {\overset{ˉ}{z}}_{2};$ (v) agar $z_{2} \neq 0$ to $\overline{(\frac{z_{1}}{z_{2}})} = \frac{{\overset{ˉ}{z}}_{1}}{{\overset{ˉ}{z}}_{2}}$ .

$z_{1} = a + i b$ , $z_{2} = c + i d$ rakho. (iv) ke liye dono sides barabar hain: $(a \pm c) - i (b \pm d) .$ (iii) ke liye product formula (4.4) ka conjugate lo: real part $a c - b d$ waisa hi rehta hai aur imaginary part $a d + b c$ ka sign badal jaata hai. ${\overset{ˉ}{z}}_{1} = a - i b$ ko ${\overset{ˉ}{z}}_{2} = c - i d$ se seedha multiply karne par bhi wahi real part $a c - b d$ aur imaginary part $- (a d + b c)$ milta hai, isliye dono barabar hain.

(i) ke liye (4.9) aur phir (iii) use karo: $∣ z_{1} z_{2} ∣^{2} = (z_{1} z_{2}) \overline{z_{1} z_{2}} = (z_{1} {\overset{ˉ}{z}}_{1}) (z_{2} {\overset{ˉ}{z}}_{2}) = ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2} .$ Dono moduli non-negative hain, isliye square roots lo. Aakhir mein, $z_{1} = \frac{z_{1}}{z_{2}} \cdot z_{2}$ likho aur is product par (i) aur (iii) lagao; $∣ z_{2} ∣$ se ya ${\overset{ˉ}{z}}_{2}$ se divide karne par (ii) aur (v) milte hain.

Dhyaan do ki list mein kya nahi hai: modulus add nahi hota. $z_{1} = 1$ aur $z_{2} = - 1$ ke liye $∣ z_{1} + z_{2} ∣ = 0$ hai lekin $∣ z_{1} ∣ + ∣ z_{2} ∣ = 2$ . General case mein sirf $∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$ sahi hai, jise triangle inequality kehte hain, aur ise tum Fig. 4.11 ke sum mode mein dekh sakte ho.

Example 4.5NCERT Example 5 · p. 82

Ek multiplicative inverse

$- 3 + 4 i$ ka multiplicative inverse nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.6NCERT Example 6 · p. 82

Conjugate se divide karna

$a + i b$ ki form mein express karo: (i) $\frac{3 + i \sqrt{3}}{1 - i \sqrt{3}}$ (ii) $i^{- 45}$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 8 questions

Practice 4.1: answers ke saath

Har answer $a + i b$ ki form mein likho, jab tak question kuch aur na maange.

1
Short answerEasyNCERT Exercise 4.1, Q1–Q3
$a + i b$ ki form mein express karo.
1. $(- 4 i) (\frac{3}{8} i)$
  Solution dikhao
  Answer $\frac{3}{2} + 0 i$
  $(- 4 i) (\frac{3}{8} i) = - \frac{12}{8} i^{2} = \frac{3}{2} .$
2. $i^{14} + i^{27}$
  Solution dikhao
  Answer $- 1 - i$
  $14 = 4 \cdot 3 + 2$ aur $27 = 4 \cdot 6 + 3$ , isliye sum hai $i^{2} + i^{3} = - 1 - i .$
3. $i^{- 71}$
  Solution dikhao
  Answer $0 + i$
  $- 71 = 4 (- 18) + 1,$ isliye $i^{- 71} = i$ . Check: $i^{71} = i^{3} = - i$ aur $\frac{1}{- i} = i$ .
2
Short answerEasyNCERT Exercise 4.1, Q4–Q7
Simplify karke $a + i b$ ki form mein likho.
1. $2 (5 - 3 i) - i (5 - 3 i)$
  Solution dikhao
  Answer $7 - 11 i$
  $10 - 6 i - 5 i + 3 i^{2} = 10 - 3 - 11 i = 7 - 11 i .$
2. $(- 2 + 5 i) - (3 - 4 i)$
  Solution dikhao
  Answer $- 5 + 9 i$
  Part by part subtract karo: $(- 2 - 3) + (5 + 4) i = - 5 + 9 i .$
3. $(\frac{3}{4} - \frac{1}{2} i) - (2 + \frac{5}{3} i)$
  Solution dikhao
  Answer $- \frac{5}{4} - \frac{13}{6} i$
  Real part hai $\frac{3}{4} - 2 = - \frac{5}{4}$ . Imaginary part hai $- \frac{1}{2} - \frac{5}{3} = - \frac{13}{6} .$
4. $[(\frac{1}{2} + \frac{3}{2} i) + (3 - \frac{1}{2} i)] - (- \frac{5}{2} + 2 i)$
  Solution dikhao
  Answer $6 - i$
  Square bracket $\frac{7}{2} + i$ hai. Subtract karne par milta hai $(\frac{7}{2} + \frac{5}{2}) + (1 - 2) i = 6 - i .$
3
Short answerMediumNCERT Exercise 4.1, Q8–Q10
Expand karke $a + i b$ ki form mein likho.
1. $(1 + i)^{6}$
  Solution dikhao
  Answer $0 - 8 i$
  $(1 + i)^{2} = 1 + 2 i + i^{2} = 2 i,$ isliye $(1 + i)^{6} = (2 i)^{3} = 8 i^{3} = - 8 i .$
2. ${(\frac{1}{2} - 2 i)}^{3}$
  Solution dikhao
  Answer $- \frac{47}{8} + \frac{13}{2} i$
  (4.7) se $(z_{1} - z_{2})^{3}$ ko $z_{1} = \frac{1}{2}$ aur $z_{2} = 2 i$ ke saath use karo; dhyaan do ki $(2 i)^{2} = - 4$ aur $(2 i)^{3} = - 8 i$ .
  $\begin{aligned} \frac{1}{8} - \frac{3}{4} (2 i) + \frac{3}{2} (2 i)^{2} - (2 i)^{3} \\ = \frac{1}{8} - \frac{3}{2} i - 6 + 8 i \\ = - \frac{47}{8} + \frac{13}{2} i \end{aligned}$
3. ${(- 1 + \frac{1}{2} i)}^{3}$
  Solution dikhao
  Answer $- \frac{1}{4} + \frac{11}{8} i$
  Pehle square karo: ${(- 1 + \frac{1}{2} i)}^{2} = 1 - i - \frac{1}{4} = \frac{3}{4} - i .$
  $\begin{aligned} (\frac{3}{4} - i) (- 1 + \frac{1}{2} i) \\ = - \frac{3}{4} + \frac{3}{8} i + i + \frac{1}{2} \\ = - \frac{1}{4} + \frac{11}{8} i \end{aligned}$
4
Short answerEasyNCERT Exercise 4.1, Q11–Q13
Har number ka multiplicative inverse nikalo.
1. $3 - 5 i$
  Solution dikhao
  Answer $\frac{3}{34} + \frac{5}{34} i$
  $∣ z ∣^{2} = 9 + 25 = 34$ aur $\overset{ˉ}{z} = 3 + 5 i$ , isliye $z^{- 1} = \frac{3 + 5 i}{34} .$
2. $\sqrt{7} - 3 i$
  Solution dikhao
  Answer $\frac{\sqrt{7}}{16} + \frac{3}{16} i$
  $∣ z ∣^{2} = 7 + 9 = 16$ , isliye $z^{- 1} = \frac{\sqrt{7} + 3 i}{16} .$
3. $2 i$
  Solution dikhao
  Answer $0 - \frac{1}{2} i$
  $\overset{ˉ}{z} = - 2 i$ aur $∣ z ∣^{2} = 4$ , isliye $z^{- 1} = - \frac{1}{2} i$ . Check: $2 i \cdot (- \frac{1}{2} i) = - i^{2} = 1.$
Short answerHardNCERT Exercise 4.1, Q14
$\frac{(\sqrt{3} + 2 i) (\sqrt{3} - 2 i)}{(1 + i \sqrt{2})^{2}}$ ko $a + i b$ ki form mein express karo.
Solution dikhao
Answer $- \frac{7}{9} - \frac{14 \sqrt{2}}{9} i$
Numerator: $∣ \sqrt{3} + 2 i ∣^{2} = 3 + 4 = 7.$ Denominator: $(1 + i \sqrt{2})^{2} = 1 + 2 i \sqrt{2} - 2 = - 1 + 2 \sqrt{2} i .$
Upar aur neeche dono ko $- 1 - 2 \sqrt{2} i$ se multiply karo; naya denominator $1 + 8 = 9$ hai.
$\begin{array}{c} \frac{7 (- 1 - 2 \sqrt{2} i)}{9} \\ = - \frac{7}{9} - \frac{14 \sqrt{2}}{9} i \end{array}$
Quick answerEasy
$i^{2026}$ kiske barabar hai?
$1$ $- 1$ $i$ $- i$
Answer dekho
Answer (B) $- 1$
$2026 = 4 \times 506 + 2,$ isliye $i^{2026} = i^{2} = - 1$ .
Quick answerEasy
$\frac{1}{1 - i}$ kiske barabar hai?
$1 + i$ $\frac{1}{2} - \frac{1}{2} i$ $\frac{1}{2} + \frac{1}{2} i$ $- 1 + i$
Answer dekho
Answer (C) $\frac{1}{2} + \frac{1}{2} i$
$\frac{1 + i}{(1 - i) (1 + i)} = \frac{1 + i}{2} .$ Option B answer ka conjugate hai.
Quick answerMedium
Nikalo: $Re (\frac{5}{2 - i}) .$
Answer dekho
Answer 2
$\frac{5}{2 - i} = \frac{5 (2 + i)}{4 + 1} = 2 + i,$ isliye real part $2$ hai.

4.5 Argand Plane aur Polar Representation

Complex number $a + i b$ real numbers ke ordered pair $(a, b)$ se poori tarah fix ho jaata hai, aur ordered pair hi to coordinate plane ke kisi point ka naam hota hai. Isliye hum point $(a, b)$ se number $a + i b$ ko represent karte hain. Har complex number ko apna ek point milta hai, aur har point ko apna ek complex number.

Aise plane ke do naam hain, complex plane aur Argand plane (आर्गंड तल); doosra naam Switzerland mein paida hue amateur mathematician Jean-Robert Argand ke naam par hai, jinhone 1806 mein ise describe kiya.Norwegian-Danish surveyor Caspar Wessel ne yahi idea 1799 mein publish kar diya tha, lekin unke paper par lagbhag ek sadi tak kisi ka dhyaan nahi gaya. Horizontal axis par numbers $a + 0 i$ hote hain aur ise real axis kehte hain; vertical axis par numbers $0 + i b$ hote hain aur yeh imaginary axis hai (Fig. 4.7).

3/3 $a + i b$ tak pahunchne ke liye $a$ across aur $b$ upar jaao, bilkul coordinate pair ki tarah.

Fig. 4.7Point

(a, b)

complex number

a + i b

ko represent karta hai.NCERT Fig. 4.1 · p. 83

Ek grid jismein horizontal axis ka label Re hai, yaani real axis, aur vertical axis ka label Im, yaani imaginary axis, jo O par cross karte hain. Step 1 real numbers −3 aur 4 ko real axis par gol dots ki tarah mark karta hai. Step 2 purely imaginary numbers 2i aur −3i ko imaginary axis par square dots ki tarah jodta hai. Step 3 chaar diamond-shaped points jodta hai: first quadrant mein 2 + 3i, second mein −4 + 2i, third mein −3 − 2i aur fourth mein 4 − 2i, saath mein 2 + 3i se neeche real axis tak aur across imaginary axis tak dashed lines.

Point padhna coordinates padhne jaisa hai: real axis ke along $a$ units chalo, phir imaginary axis ke parallel $b$ units. Number $2 + 3 i$ first quadrant mein point $(2, 3)$ hai, jabki $- 3 - 2 i$ third quadrant mein $(- 3, - 2)$ par hai. Axes par khud real numbers aur purely imaginary numbers hote hain.

Plane se modulus ko ek matlab milta hai. $z = a + i b$ ka point origin O se horizontal distance $∣ a ∣$ aur vertical distance $∣ b ∣$ par hai, isliye Pythagoras se O se uski distance $\sqrt{a^{2} + b^{2}} = ∣ z ∣$ hai (Fig. 4.8). Kai numbers ka modulus same hota hai: $∣ z ∣ = 5$ wale saare numbers O ke around radius $5$ ke circle par hote hain.

2/3Legs $4$ aur $3$ hain, isliye $∣ z ∣ = \sqrt{4^{2} + 3^{2}} = 5.$

Fig. 4.8

∣ z ∣

point

z

ki O se distance hai.NCERT Fig. 4.2 · p. 84

Ek Argand plane jismein point z = 4 + 3i hai. Step 2 dashed legs banata hai, real axis ke along length 4 aur upar ki taraf 3, right-angle mark ke saath; O se z tak ke segment ka label |z| = 5 hai, aur working root of 4 squared plus 3 squared = 5 likhi hai. Step 3 O ke around radius 5 ka dotted circle jodta hai jo −3 + 4i, −5, −5i aur 3 − 4i se guzarta hai, aur yeh sab points mark kiye gaye hain.

Conjugate ek reflection ban jaata hai. $z$ ka point $(a, b)$ aur $\overset{ˉ}{z}$ ka point $(a, - b)$ ek hi vertical line par hain, real axis se barabar distance upar aur neeche. Isliye $\overset{ˉ}{z}$ real axis mein $z$ ki mirror image hai (Fig. 4.9). Isi tarah, $- z$ O ke around half-turn ke baad $z$ ki image hai, aur $- \overset{ˉ}{z}$ imaginary axis mein uski mirror image hai.

3/3 $- z$ O ke around half-turn wali image hai; $- \overset{ˉ}{z}$ imaginary axis mein mirror image hai.

Fig. 4.9

\overset{ˉ}{z}

real axis mein

z

ki mirror image hai.NCERT Fig. 4.3 · p. 84

Ek Argand plane jismein point z = 3 + 2i hai. Step 2 real axis ko highlight karta hai, z se neeche conjugate 3 − 2i tak dashed vertical segment banata hai, jahan woh axis ko cross karta hai wahan right-angle mark aur uske dono halves par barabar tick marks ke saath. Step 3 points −z = −3 − 2i, jo O se hoti hui dashed line se z se juda hai, aur −z-bar = −3 + 2i jodta hai, aur chaaron points se hota hua ek dotted rectangle banata hai.

Example

Argand plane par distances

$z = - 4 + 3 i$ ke liye $∣ z ∣$ nikalo, $\overset{ˉ}{z}$ aur $- z$ plot karo, aur $z$ se dono ki distance nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

SimulationPlane par modulus aur conjugate

Dono axes par −6 se 6 tak ek square grid, origin par fixed point O aur ek draggable point z jo (3, 2), yaani 3 + 2i, se shuru hota hai. O se z tak segment bana hai aur uski length measure hoti hai, jo z ka modulus hai. Har point ki horizontal axis mein mirror image bhi bani hai, jo z ke liye uska conjugate hai. z ko whole-number points ke beech drag karne par coordinates update hote hain.

Fig. 4.10Horizontal axis ko real axis aur vertical axis ko imaginary axis samjho.

z

ko drag karo: measured length

∣ z ∣

hai aur mirror image

\overset{ˉ}{z}

hai.

Try karo

$z$ ko $(4, 3)$ par drag karo aur distance padho. Phir O se itni hi distance par teen aur grid points dhoondho.
$z$ ko horizontal axis par rakho. Uski mirror image kahan hai, aur isse real number ke conjugate ke baare mein kya pata chalta hai?
$z$ ko third quadrant mein le jaao. $\overset{ˉ}{z}$ kis quadrant mein hai?

InteractiveComplex arithmetic ki lab

Dono axes par −8 se 8 tak ek square grid, real axis horizontal, origin se do arrows ke saath: z1, solid blue, jo 2 + i se shuru hota hai, aur z2, dashed orange, jo 1 + 2i se shuru hota hai. Unke tips ko −5 se 5 tak ke whole-number points par drag kiya ja sakta hai ya chaar number boxes se set kiya ja sakta hai. Chhe buttons view chunte hain. Sum z1 aur z2 par bane parallelogram ko shade karta hai aur sum ko uske diagonal ki tarah banata hai. Difference z1 ke tip se minus z2 banata hai. Product z1 times z2 banata hai aur moduli compare karta hai. Quotient conjugate se multiply karta hai aur answer fraction aur decimal mein deta hai. Conjugate z1 ko real axis mein reflect karta hai aur radius |z1| ka circle banata hai. Times i z1, i z1, i squared z1 aur i cubed z1 ko ek square ke corners par dikhata hai. Ek panel har result ko words aur symbols mein batata hai. Jab tak learner kuch nahi karta, kuch nahi hilta.

Fig. 4.11

z_{1}

aur

z_{2}

ko drag karo ya unke parts type karo, phir ek operation chuno. Panel algebra dikhata hai aur grid dikhata hai ki answer kahan hai.

Quick check

3 mein se 1

- 3 - 4 i

Argand plane ke kis quadrant mein hai?

FirstSecondThirdFourth

Answer dekho

Answer (C) Third

Real part negative hai (imaginary axis ke left) aur imaginary part negative hai (real axis ke neeche), isliye point third quadrant mein hai. Second quadrant chunne ki galti sirf real part check karne se hoti hai.

Miscellaneous Examples

Example 4.7NCERT Example 7 · p. 85

Quotient ka conjugate

Iska conjugate nikalo: $\frac{(1 + 3 i) (2 - i)}{(1 - i) (3 + i)} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.8NCERT Example 8 · p. 85

Ek condition jo circle par le aati hai

Maan lo $z = x + i y$ aur $z \neq - 1$ . Agar $\frac{z - 1}{z + 1}$ purely imaginary hai, to prove karo ki $x^{2} + y^{2} = 1$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Geometrically, $∣ z ∣ = 1$ ka matlab hai ki $z$ O ke around radius $1$ ke circle par hai, jise unit circle kehte hain (Fig. 4.12). Figure kuch aur bhi dikhata hai: $z$ ko $- 1$ aur $1$ se jodne wale segments hamesha right angle par milte hain, yaani semicircle mein bana angle. Aage jab tum complex number ke argument ke baare mein padhoge, to dekhoge ki yahi right angle quotient ko purely imaginary banata hai.

3/3Dono segments $90^{\circ}$ par milte hain: semicircle mein bana angle.

Fig. 4.12

∣ z ∣ = 1

wale points se unit circle banta hai.NCERT Example 8 · p. 85

Origin O par centre wala radius 1 ka circle, jo real axis ko −1 aur 1 par cross karta hai. Step 1 circle dikhata hai, jiska label x squared + y squared = 1 hai, aur point z = 0.6 + 0.8i tak length 1 ka dashed radius. Step 2 z ko −1 se solid segment se aur 1 se dashed segment se jodta hai. Step 3 radius ko fade karta hai aur z par dono segments ke beech right angle mark karta hai, label 90 degrees.

Recap

Summary aur key points

Real numbers se $x^{2} + 1 = 0$ solve nahi hota; $i^{2} = - 1$ wala $i$ jodne se woh complex numbers tak extend ho jaate hain, aur $𝐑 \subset 𝐂$ (Section 4.1).
Complex number $z = a + i b$ hota hai, jiska real part $a$ aur imaginary part $b$ hai, dono real; do complex numbers tab barabar hote hain jab dono parts match karein (Section 4.2).
$(a + i b) \pm (c + i d) = (a \pm c) + i (b \pm d)$ (Section 4.3.1).
$(a + i b) (c + i d) = (a c - b d) + i (a d + b c) .$ Addition aur multiplication closed, commutative aur associative hain, inki identities $0$ aur $1$ hain, inverses hote hain, aur yeh distributive law follow karte hain (Section 4.3.3).
Har $z = a + i b \neq 0$ ka inverse hota hai $z^{- 1} = \frac{a - i b}{a^{2} + b^{2}},$ aur $\frac{z_{1}}{z_{2}} = z_{1} \cdot \frac{1}{z_{2}}$ (Section 4.3.4).
Har integer $k$ ke liye $i^{4 k} = 1$ , $i^{4 k + 1} = i$ , $i^{4 k + 2} = - 1$ aur $i^{4 k + 3} = - i$ (Section 4.3.5).
$a > 0$ ke liye $\sqrt{- a} = i \sqrt{a}$ ; jab $a$ aur $b$ dono negative hon, to $\sqrt{a} \sqrt{b}$ ko kabhi $\sqrt{a b}$ mein merge mat karo (Section 4.3.6).
Real numbers ki identities, jaise $(z_{1} + z_{2})^{2} = z_{1}^{2} + 2 z_{1} z_{2} + z_{2}^{2},$ complex numbers ke liye bhi sahi hain (Section 4.3.7).
$∣ z ∣ = \sqrt{a^{2} + b^{2}}$ , $\overset{ˉ}{z} = a - i b$ , $z \overset{ˉ}{z} = ∣ z ∣^{2}$ aur $z^{- 1} = \overset{ˉ}{z} / ∣ z ∣^{2}$ ; moduli multiply aur divide hote hain, aur conjugate chaaron operations ke saath andar chala jaata hai (Section 4.4).
Argand plane mein $z$ point $(a, b)$ hai, $∣ z ∣$ O se uski distance hai aur $\overset{ˉ}{z}$ real axis mein uski mirror image hai (Section 4.5).

Timeline

Impossible roots kaise numbers bane

c. 850 CE
Mahavira
Ganita Sara Sangraha mein woh likhte hain ki kisi bhi number ka square negative nahi hota, isliye jaane-pehchaane numbers mein negatives ke square roots nahi hain.
1150
Bhaskara II
Unka Bijaganita negative quantities ke baare mein yahi baat kehta hai.
1545
Gerolamo Cardano
$10$ ko aise do parts mein baantne se jinka product $40$ ho, woh $5 \pm \sqrt{- 15}$ tak pahunchte hain, jise woh bekaar keh kar chhod dete hain.
1629
Albert Girard
Unka argument hai ki aise roots maan lene se equation ke utne hi roots ho sakte hain jitni uski degree hai.
1777
Leonhard Euler
Woh $\sqrt{- 1}$ ke liye letter $i$ introduce karte hain.
1799–1806
Wessel aur Argand
Dono complex numbers ko ek plane ke points ki tarah dekhte hain.
1830s
W. R. Hamilton
Woh $a + i b$ ko real numbers ke ordered pair $(a, b)$ ki tarah define karte hain, add aur multiply karne ke rules ke saath, aur saara rahasya khatam kar dete hain.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 4.1, Q1–Q3	Practice 4.1 · Q1
NCERT Exercise 4.1, Q4–Q7	Practice 4.1 · Q2
NCERT Exercise 4.1, Q8–Q10	Practice 4.1 · Q3
NCERT Exercise 4.1, Q11–Q13	Practice 4.1 · Q4
NCERT Exercise 4.1, Q14	Practice 4.1 · Q5
NCERT Miscellaneous Exercise, Q1	Chapter 4 par miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Chapter 4 par miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Chapter 4 par miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Chapter 4 par miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Chapter 4 par miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Chapter 4 par miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Chapter 4 par miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8	Chapter 4 par miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9	Chapter 4 par miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10	Chapter 4 par miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q11	Chapter 4 par miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q12	Chapter 4 par miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q13	Chapter 4 par miscellaneous exercises · Q13
NCERT Miscellaneous Exercise, Q14	Chapter 4 par miscellaneous exercises · Q14

Quick check: MCQs

4.2 Complex Numbers3 questions
4.3.4 Do complex numbers ka division3 questions
4.3.7 Identities3 questions
4.4 Complex Number ka Modulus aur Conjugate4 questions
4.5 Argand Plane aur Polar Representation3 questions
Miscellaneous Examples3 questions

Glossary · 12

Key terms (English + Hindi)

additive inverseयोज्य प्रतिलोमyojya pratilom: $z = a + i b$ ke liye number $- z = - a - i b$ ; dono ka sum $0$ hota hai.Chapter mein pehli baar yahan aaya →
Argand planeआर्गंड तलargand tal: Coordinate plane jismein point $(a, b)$ complex number $a + i b$ ko represent karta hai.Chapter mein pehli baar yahan aaya →

complex numberसम्मिश्र संख्याsammishra sankhya: Number $a + i b$ jismein $a$ aur $b$ real hain aur $i^{2} = - 1$ .Chapter mein pehli baar yahan aaya →
conjugateसंयुग्मीsanyugmi: $z = a + i b$ ke liye number $\overset{ˉ}{z} = a - i b$ , jismein imaginary part ka sign ulta hai.Chapter mein pehli baar yahan aaya →

imaginary axisकाल्पनिक अक्षkalpanik aksh: Argand plane ka vertical axis; iske points numbers $0 + i b$ hain.Chapter mein pehli baar yahan aaya →
imaginary partकाल्पनिक भागkalpanik bhag: $z = a + i b$ ke liye real number $b$ ( $i b$ nahi), jise $Im z$ likhte hain.Chapter mein pehli baar yahan aaya →
imaginary unitकाल्पनिक इकाईkalpanik ikai: Number $i$ jiska square $- 1$ hai. Ise $\sqrt{- 1}$ likhte hain aur 'iota' padhte hain.Chapter mein pehli baar yahan aaya →

modulusमापांकmaapaank: $z = a + i b$ ke liye non-negative real number $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ , yaani origin se $z$ ki distance.Chapter mein pehli baar yahan aaya →
multiplicative inverseगुणात्मक प्रतिलोमgunatmak pratilom: $z \neq 0$ ke liye number $z^{- 1}$ jismein $z z^{- 1} = 1$ . Yeh $\overset{ˉ}{z} / ∣ z ∣^{2}$ ke barabar hai.Chapter mein pehli baar yahan aaya →

purely imaginary numberशुद्ध काल्पनिक संख्याshuddh kalpanik sankhya: Complex number $0 + i b$ jismein $b \neq 0$ , jaise $7 i$ .Chapter mein pehli baar yahan aaya →

real axisवास्तविक अक्षvastavik aksh: Argand plane ka horizontal axis; iske points real numbers $a + 0 i$ hain.Chapter mein pehli baar yahan aaya →
real partवास्तविक भागvastavik bhag: $z = a + i b$ ke liye real number $a$ , jise $Re z$ likhte hain.Chapter mein pehli baar yahan aaya →

Complex Numbers and Quadratic Equations: Class 11 Maths Chapter 4 Hinglish mein

Is chapter mein aap kya seekhenge

4.1 Introduction

4.2 Complex Numbers

4.3 Complex Numbers ka Algebra

4.3.1 Do complex numbers ka addition

4.3.2 Do complex numbers ka difference

4.3.3 Do complex numbers ka multiplication

4.3.4 Do complex numbers ka division

4.3.5 i ki Power

4.3.6 Negative real number ke square roots

4.3.7 Identities

4.4 Complex Number ka Modulus aur Conjugate

Practice 4.1: answers ke saath

4.5 Argand Plane aur Polar Representation

Miscellaneous Examples

Practice questions answers ke saath (NCERT “Chapter 4 par miscellaneous exercises”)

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye