Applications — solving polynomial equations with complex roots

3.5.13 · Maths › Complex Numbers

1. The starting block: what is a root?

KYA chahiye hume: saare $α_{k}$ , chahe kuch complex hi kyun na hon.

2. The Conjugate Root Theorem (the workhorse)

ISKO DERIVE KAISE KARTE HAIN (scratch se). Conjugation ke baare mein do facts use karte hain: $\overline{z + w} = \overset{z}{ˉ} + \overset{w}{ˉ}, \overline{z w} = \overset{z}{ˉ} \overset{w}{ˉ} \Rightarrow \overline{z^{k}} = \overset{z}{ˉ}^{k} .$

Poori equation $P (α) = 0$ ka conjugate lete hain: $\overline{P (α)} = \overline{\sum_{k} a_{k} α^{k}} = \sum_{k} \overline{a_{k}} \overline{α^{k}} .$

Yeh step kyun? Conjugation sums aur products ke upar distribute hoti hai, isliye hum bar ko andar push kar sakte hain.

Kyunki coefficients real hain, $\overline{a_{k}} = a_{k}$ . Aur $\overline{α^{k}} = \overset{α}{ˉ}^{k}$ . Toh $\overline{P (α)} = \sum_{k} a_{k} \overset{α}{ˉ}^{k} = P (\overset{α}{ˉ}) .$

Lekin $\overline{P (α)} = \overline{0} = 0$ . Isliye $P (\overset{α}{ˉ}) = 0$ : $\overset{α}{ˉ}$ ek root hai. $■$

3. Strategy toolbox (the 80/20 core)

Yeh teen techniques almost sabkuch solve kar deti hain:

Tool	Kab use karein	Idea
Quadratic formula	degree 2	$z = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , $negative$ allow karo
Conjugate pairs + factoring	ek complex root diya gaya ho	use $\overset{α}{ˉ}$ se pair karo, real quadratic divide karo
De Moivre / roots of unity	$z^{n} = c$	ek circle par $n$ equally-spaced points

Figure — Applications — solving polynomial equations with complex roots

4. Quadratics with negative discriminant

KAISE. $a z^{2} + b z + c = 0$ par complete the square karo: $a (z + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a} .$

Yeh step kyun? Complete the square se squared term isolate ho jaati hai taaki hum root le sakein.

Jab $b^{2} - 4 a c < 0$ ho, toh right side negative hoti hai, isliye square root lene par $i$ aata hai: $z + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a} = \pm \frac{i 4 a c - b ^{2}}{2 a} .$

5. Ek diya gaya complex root use karke ek higher polynomial crack karna

6. $z^{n} = c$ ke Roots (De Moivre)

KAISE. Agar $z = ρ (cos ϕ + i sin ϕ)$ ho toh De Moivre se $z^{n} = ρ^{n} (cos n ϕ + i sin n ϕ)$ . $c$ se match karo: modulus deta hai $ρ^{n} = r \Rightarrow ρ = r^{1/ n}$ ; angle deta hai $n ϕ = θ + 2 π k$ (kyunki poore turns add karne se point nahi badlta). $ϕ$ ke liye solve karo.

$+ 2 π k$ kyun? Kyunki $θ$ aur $θ + 2 π$ same direction hain, lekin $n$ se divide karne par woh $n$ distinct angles mein spread ho jaate hain.

7. Vieta sanity-checks (free verification)

Flashcards

Fundamental Theorem of Algebra kya guarantee karta hai?

Har degree-

n

polynomial ke exactly

n

complex roots hote hain, multiplicity count karte hue.

Conjugate Root Theorem batao.

Agar ek polynomial ke real coefficients hain aur

p + q i

ek root hai, toh

p - q i

bhi ek root hai.

Ek real odd-degree polynomial mein kam se kam ek real root kyun hoti hai?

Complex roots conjugate pairs mein aati hain (even count), toh odd wala bacha hua real hona chahiye.

z^{2} - 4 z + 13 = 0

ke roots?

2 \pm 3 i

Roots

2 \pm 3 i

se real quadratic factor kya hoga?

z^{2} - 4 z + 13

(sum

4

, product

13

z^{n} = c

n

solutions ka formula?

z_{k} = r^{1/ n} (cos \frac{θ + 2 π k}{n} + i sin \frac{θ + 2 π k}{n}), k = 0.. n - 1

Vieta:

a_{n} z^{n} + \dots + a_{0}

ke roots ka sum aur product?

Sum

= - a_{n - 1} / a_{n}

; product

= (- 1)^{n} a_{0} / a_{n}

Ek quartic reduce karne ke liye ek diya gaya complex root kaise use karte hain?

Use apne conjugate ke saath pair karo taaki ek real quadratic factor mile, use divide karo, bacha hua quadratic solve karo.

Recall Feynman: ek 12-saal ke bacche ko explain karo

Number line par tum koi aisa number nahi dhundh sakte jiska square $- 1$ ho — squaring hamesha positive deta hai. Toh mathematicians ne ek aisa number imagine kiya, jise $i$ kehte hain, jo line ke upar rehta hai. Ab har "impossible" equation ke answers hain; woh bas thode side mein ek flat plane mein dots ki tarah baithe hain. Aur ek cool trick hai: agar koi equation sirf ordinary (real) numbers se bani ho, toh uske hidden answers hamesha mirror-image twins ki tarah aate hain — ek upar, ek neeche line ke. Ek twin dhundho aur tum turant doosra jaante ho.

Connections

Complex Numbers — Conjugate and Modulus (proof mein use ki gayi $\overset{z}{ˉ}$ ki algebra)
De Moivre's Theorem ( $z^{n} = c$ ke roots)
Roots of Unity (special case $c = 1$ )
Vieta's Formulas (sum/product checks)
Fundamental Theorem of Algebra
Polynomial Division (known factors divide karna)