De Moivre's theorem — statement, proof, applications

3.5.10 · Maths › Complex Numbers

WHAT is the statement?

WHY does it look this way? Kyunki $cos θ + i sin θ = e^{i θ}$ (Euler), toh LHS ban jaata hai $(e^{i θ})^{n} = e^{in θ} = cos n θ + i sin n θ$ . Yeh "cheat" proof hai — lekin hume ise scratch se earn karna chahiye.

HOW to prove it (from first principles)

Step 0 — the key multiplication fact

Maano $z_{1} = cis α$ aur $z_{2} = cis β$ . Multiply karo: $z_{1} z_{2} = (cos α + i sin α) (cos β + i sin β) .$ Why expand? Real aur imaginary parts collect karne ke liye. $= (cos α cos β - sin α sin β) + i (sin α cos β + cos α sin β) .$ Why does this simplify? Yeh do brackets exactly angle-addition formulas hain: $= cos (α + β) + i sin (α + β) = cis (α + β) .$ Toh cis's ko multiply karne par angles add ho jaate hain. Yeh ek fact hi sab kuch chalata hai.

Step 1 — positive integers by induction

Base case $n = 1$ : $(cis θ)^{1} = cis θ = cis (1 \cdot θ)$ . ✔

Inductive step: maan lo $(cis θ)^{k} = cis (k θ)$ . Tab $(cis θ)^{k + 1} = (cis θ)^{k} \cdot cis θ = cis (k θ) \cdot cis θ .$ Why can I use Step 0 now? Kyunki hum do cis's ko multiply kar rahe hain — toh angles add ho jaayenge: $= cis (k θ + θ) = cis ((k + 1) θ) . ✓$ Induction se yeh sabhi positive integers ke liye hold karta hai.

Step 2 — $n = 0$

$(cis θ)^{0} = 1 = cos 0 + i sin 0 = cis (0 \cdot θ)$ . ✔

Step 3 — negative integers

Maano $n = - m$ jahan $m > 0$ hai. Tab $(cis θ)^{- m} = \frac{1}{( cis θ ) ^{m}} = \frac{1}{cis ( m θ )}$ . Why is the reciprocal easy? Kyunki $∣ cis (m θ) ∣ = 1$ , isliye uska inverse uska conjugate hai: $\frac{1}{c o s m θ + i s i n m θ} = \frac{c o s m θ - i s i n m θ}{c o s ^{2} m θ + s i n ^{2} m θ} = cos m θ - i sin m θ = cis (- m θ) .$ Toh $(cis θ)^{- m} = cis (- m θ) = cis (n θ)$ . ✔ Theorem sabhi integers ke liye hold karta hai.

Applications

A) $n$ -th roots of a complex number

$z^{n} = w$ solve karne ke liye jahan $w = r cis ϕ$ : likho $z = ρ cis ψ$ . Tab $z^{n} = ρ^{n} cis (n ψ) = r cis ϕ$ . Why add $2 π k$ ? Kyunki angles $ϕ$ aur $ϕ + 2 π k$ same $w$ ko label karte hain. Match karke: $ρ = r^{1/ n}, n ψ = ϕ + 2 π k ⟹ ψ = \frac{ϕ + 2 π k}{n} .$ $z_{k} = r^{1/ n} cis (\frac{ϕ + 2 π k}{n}), k = 0, 1, \dots, n - 1.$ Yeh $n$ points hain jo radius $r^{1/ n}$ wale circle par $2 π / n$ se equally spaced hain.

B) Multiple-angle identities

$(cos θ + i sin θ)^{n}$ ko binomial theorem se expand karo, phir real/imaginary parts ko $cos n θ$ aur $sin n θ$ ke saath equate karo.

C) Sums like $\sum cos k θ$ via geometric series of $cis k θ$ .

Worked examples

Recall Feynman: explain to a 12-year-old

Ek clock ki sui ki kalpana karo jiska kuch length hai. Har baar jab tum us sui ko "multiply" karte ho khud se, woh apne angle se ghoomti hai aur stretch hoti hai. Yeh $n$ baar karo aur sui ne apna angle $n$ baar ghoom liya hoga aur $n$ guna stretch ho gayi hogi. Yahi poora theorem hai: ghoomna add hota hai, stretching multiply hoti hai. Ulta jaana (roots dhundna) matlab: $n$ steps mein kahin pahunchne ke liye, angle ko $n$ equal slices mein kaato — aur kyunki ek poora chakkar ( $360°$ ) tumhe wapas wahi laata hai, circle ke around exactly $n$ starting spots hain equally spaced.

Flashcards

De Moivre's theorem statement (integer

n

)

(cos θ + i sin θ)^{n} = cos n θ + i sin n θ

Proof ke multiplication step mein use ki gayi key identity

The angle-addition formulas:

cis α \cdot cis β = cis (α + β)

n

-th power ke under modulus aur argument ka kya hota hai?

Modulus

\to r^{n}

(multiply hota hai), argument

\to n θ

(add hota hai).

r cis ϕ

n

n

-th roots ka formula

z_{k} = r^{1/ n} cis (\frac{ϕ + 2 π k}{n}), k = 0, \dots, n - 1

Exactly

n

roots kyun?

Argument mein

2 π k

add karne par sirf

k = 0, \dots, n - 1

ke liye distinct angles milte hain (phir repeat ho jaate hain).

cos 3 θ

in terms of

cos θ

4 cos^{3} θ - 3 cos θ

sin 3 θ

in terms of

sin θ

3 sin θ - 4 sin^{3} θ

Negative

n

ke liye De Moivre prove kaise karein?

Reciprocal lo; kyunki

∣ cis (m θ) ∣ = 1

, uska inverse uska conjugate

cis (- m θ)

hai.

(1 + i)^{8}

ki value

16

Unity ke teen cube roots ka sum

0

Connections

Polar form of complex numbers — $r cis θ$ representation jis par yeh depend karta hai.
Euler's formula — $e^{i θ} = cis θ$ De Moivre ko one-liner bana deta hai.
Roots of unity — sabse clean application (equally spaced points).
Binomial theorem — multiple-angle identities nikalne ke liye use hota hai.
Argand diagram — spin-and-stretch ka geometric picture.
Trigonometric identities — angle-addition formulas proof ko power dete hain.