3.5.11 · Maths › Complex Numbers
n -th root lena essentially n -th power ka reverse hai. Powers angles ko n se multiply karte hain , isliye roots ko angles ko n se divide karna padta hai. Lekin ek akela complex number bahut saare equivalent angles (θ , θ + 2 π , θ + 4 π , … ) ke saath likha ja sakta hai kyunki ek poora chakkar laga ke aap wapas wahin aate ho jahan se shuru kiya tha. Jab hum inhe n se divide karte hain, ye sab equal nahi rehte — ye ==exactly n distinct roots== mein fan out ho jaate hain, ek circle ke aas-paas evenly spaced. Yahi "ek number, kai angles" wali baat poora secret hai.
Definition Complex number ka nth root
Ek complex number w aur integer n ≥ 1 diya ho, to w ka n -th root koi bhi z hai jo satisfy kare
z n = w .
Aisa exactly n z hote hain (jab w = 0 ), inhe w ke n -th roots kehte hain.
Special case w = 1 se milte hain n -th roots of unity (z n = 1 ke solutions).
Step 1 — w ko polar form mein likho.
Koi bhi nonzero w = r ( cos θ + i sin θ ) , jahan r = ∣ w ∣ > 0 aur θ = arg w .
Ye step kyun? Complex numbers ka multiplication polar form mein sabse easy hai: moduli multiply hote hain, arguments add hote hain. Roots, multiplication ko undo karne ke baare mein hain, isliye polar natural language hai.
Step 2 — Unknown root ko bhi polar form mein likho.
Maano z = ρ ( cos ϕ + i sin ϕ ) , jahan unknown modulus ρ > 0 aur unknown angle ϕ hai.
Step 3 — z n par De Moivre apply karo.
z n = ρ n ( cos n ϕ + i sin n ϕ ) .
Ye step kyun? De Moivre's theorem ( cos ϕ + i sin ϕ ) n = cos n ϕ + i sin n ϕ batata hai ki power lene se angle n se multiply hota hai aur modulus power ho jaata hai.
Step 4 — w ke saath match karo. Hume chahiye z n = w , to
ρ n ( cos n ϕ + i sin n ϕ ) = r ( cos θ + i sin θ ) .
Polar form mein do complex numbers equal hote hain tab aur sirf tab jab unke moduli equal hon aur unke arguments 2 π ke multiple se differ karein:
ρ n = r , n ϕ = θ + 2 π k , k ∈ Z .
+ 2 π k kyun? YE crucial move hai. Angles sirf full turns tak defined hain; cos aur sin ka period 2 π hai. Agar hum 2 π k bhool jaayein to sirf ek root milega. Ise rakhne se baaki saare n roots milte hain.
Step 5 — Solve karo.
ρ = r 1/ n ( real positive n -th root ) , ϕ = n θ + 2 π k .
Step 6 — Kitne distinct roots hain yeh pata karo.
Jab k = 0 , 1 , 2 , … , n − 1 to hume n alag angles milte hain. k = n ke liye angle hai n θ + 2 π — k = 0 jaisi hi direction. To values period n ke saath repeat hoti hain: sirf n distinct roots hain.
Yahan r = 1 , θ = 0 , to
z k = cos n 2 π k + i sin n 2 π k = e 2 π ik / n , k = 0 , … , n − 1.
ω = e 2 π i / n likhne par, roots hain 1 , ω , ω 2 , … , ω n − 1 — ek geometric progression .
Intuition Ye zero kyun sum hote hain
Roots ek regular polygon ke vertices hain jo origin par centred hai. Symmetry se vectors cancel ho jaate hain:
1 + ω + ω 2 + ⋯ + ω n − 1 = 0 ( n ≥ 2 ) .
Algebraically: yeh ek GP hai jo sum hoti hai ω − 1 ω n − 1 = ω − 1 1 − 1 = 0 mein.
Worked example Example 1 —
i ke Square roots
z 2 = i solve karo.
Polar form: i = 1 ⋅ ( cos 2 π + i sin 2 π ) , isliye r = 1 , θ = 2 π .
Kyun? Formula mein dalne ke liye modulus aur argument chahiye.
Roots ka modulus: r 1/2 = 1 .
Angles: ϕ k = 2 π /2 + 2 π k = 4 π + π k .
k = 0 : ϕ = 4 π ⇒ z 0 = cos 4 π + i sin 4 π = 2 1 ( 1 + i ) .
k = 1 : ϕ = 4 5 π ⇒ z 1 = − 2 1 ( 1 + i ) .
Check: ( 2 1 + i ) 2 = 2 ( 1 + i ) 2 = 2 2 i = i . ✓ Dono roots ek doosre ke negatives hain (expected — square roots hamesha aisa hi hote hain).
Worked example Example 2 —
8 ke Cube roots
z 3 = 8 solve karo.
Polar: 8 = 8 ( cos 0 + i sin 0 ) , r = 8 , θ = 0 .
Roots ka modulus: 8 1/3 = 2 .
Angles: ϕ k = 3 0 + 2 π k = 3 2 π k .
k = 0 : z 0 = 2 ( cos 0 + i sin 0 ) = 2.
k = 1 : z 1 = 2 ( cos 3 2 π + i sin 3 2 π ) = 2 ( − 2 1 + i 2 3 ) = − 1 + i 3 .
k = 2 : z 2 = 2 ( cos 3 4 π + i sin 3 4 π ) = − 1 − i 3 .
Ek real number ke liye teen roots kyun? Bhale hi 8 real hai, ek cubic z 3 − 8 = 0 ke 3 roots hote hain; sirf ek real hai, baaki do complex conjugates hain. Real answer ≠ sirf ek root.
Worked example Example 3 —
− 16 ke saare 4th roots
z 4 = − 16 solve karo.
Polar: − 16 = 16 ( cos π + i sin π ) , r = 16 , θ = π .
Modulus: 1 6 1/4 = 2 .
Angles: ϕ k = 4 π + 2 π k for k = 0 , 1 , 2 , 3 :
4 π , 4 3 π , 4 5 π , 4 7 π .
Roots: 2 cis 4 π = 2 + i 2 , phir − 2 + i 2 , − 2 − i 2 , 2 − i 2 .
Ek perfect square 45° rotate hua, radius 2 . Evenly spaced kyun? k mein har step 4 2 π = 90° add karta hai.
n -th root sirf ek number hota hai."
Ye sahi kyun lagta hai: Positive reals ke liye hum kehte hain "9 = 3 " — ek value. To log ek hi answer expect karte hain.
The fix: Yeh principal root convention hai. C par equation z n = w mein genuinely n solutions hote hain. Hamesha k = 0 , … , n − 1 tak chalao.
Common mistake Argument mein
+ 2 π k bhool jaana.
Ye sahi kyun lagta hai: arg w ek single fixed number lagta hai.
The fix: Arguments "2 π modulo" mein hote hain. 2 π k drop karne se saare roots ek mein collapse ho jaate hain. n se divide karne se pehle ise add karo.
r 1/ n kisi negative radius ka, ya r ki jagah θ ka lena.
Ye sahi kyun lagta hai: Confuse ho jaate hain ki modulus kaunsa quantity hai.
The fix: r = ∣ w ∣ hamesha ≥ 0 hota hai; aap sirf modulus ka ordinary positive real n -th root lete ho. Angle θ divide hota hai, root nahi.
Common mistake Degrees aur radians ko inconsistently use karna.
Fix: Agar θ radians mein hai to 2 π k add karo; agar degrees mein hai to 360° k add karo. Kabhi mix mat karo.
Recall Feynman: ek 12-saal ke bacche ko explain karo
Ek ghadi ki sooee imagine karo. "n ki power karna" matlab hai sooee ko n guna tez ghoomana. Ise undo karne ke liye (root lene ke liye) aapko n guna dheere ghoomana hoga. Par yahan trick hai: sooee 12 baje ki taraf point kar rahi hai — yeh 12 baje waali position ek baar poora chakkar lagate, do baar, teen baar... ke baad bhi same hi hai. Jab aap un saari "same" positions ko n se slow down karte ho, to wo n alag-alag jagah faili jaati hain, ghadi ke chakkar ke aas-paas evenly — jaise n ped ek gol taalaab ke aas-paas equally lagaaye jaayein. Saare ped centre se same doori par hain (woh doori length ka slow-down hai), bas alag-alag angles par.
Mnemonic Recipe yaad rakho:
"Same length, spread the angle."
R oot the modulus: r 1/ n .
A dd 2 π k phir D ivide by n : n θ + 2 π k .
k runs 0 to n − 1 .
"RAD-k": R oot, A dd, D ivide, run k .
#flashcards/maths
Ek nonzero complex number ke kitne distinct n -th roots hote hain? Exactly n .
z n = w jahan w = r cis θ ke k -th root ka formula kya hai?z k = r 1/ n cis ( n θ + 2 π k ) , k = 0 , … , n − 1 .
Saare n roots ka common modulus kya hota hai? r 1/ n (∣ w ∣ ka real positive n -th root).
Consecutive roots ke beech angular separation kitni hoti hai? n 2 π radians.
Argand plane mein n roots se kaunsi geometric shape banti hai? Radius r 1/ n ke circle par ek regular n -gon ke vertices.
n se divide karne se pehle 2 π k kyun add karna zaroori hai?Kyunki arg sirf mod 2 π tak defined hai; full turns add karne se baaki distinct roots milte hain.
Saare n -th roots of unity ka sum (n ≥ 2 )? 0 (symmetry / GP sum).
Roots of unity ko GP ke roop mein kaise likhte hain? 1 , ω , ω 2 , … , ω n − 1 jahan ω = e 2 π i / n .
Kisi bhi complex number ke square roots mein kya relation hota hai? Ye ek doosre ke negatives hote hain, z 1 = − z 0 .
z 3 = 8 ke liye complex roots kyun hain, jabki 8 real hai?C par ek cubic ke 3 roots hote hain; real value ≠ sirf ek root.
w = r cos theta + i sin theta
z = rho cos phi + i sin phi
z^n = rho^n cos n phi + i sin n phi
phi = theta+2 pi k over n
lie on circle radius r^1/n