KYU kaam aata hai? Yeh complex numbers ki multiplication ko angle addition mein convert karta hai, trig identities ko ek-line algebra bana deta hai, De Moivre's theorem turant de deta hai, aur maths ki sabse famous equation produce karta hai: eiπ+1=0.
KYU yeh method? Hum ex, cosx, sinx ko unki power series se define karte hain (yeh sab complex inputs ke liye converge karti hain). Phir hum bas x=iθ plug in karte hain aur i ki powers ko sorting karne dete hain.
= \sum_{n=0}^{\infty}\frac{i^n\,\theta^n}{n!}.$$
Terms likhte hain, $i^0{=}1,\ i^1{=}i,\ i^2{=}{-}1,\ i^3{=}{-}i,\dots$ use karke:
$$e^{i\theta}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \cdots$$
$$= 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots$$
*Yeh step kyun?* Har $i^n$, $+1,+i,-1,-i,\dots$ mein turn hota hai, isliye aadhe terms mein $i$ ka factor hota hai aur aadhe mein nahi.
### Step 3 — Real ($i^{even}$) aur imaginary ($i^{odd}$) terms alag karo
Woh terms **bina** $i$ ke (even $n$) aur **$i$ ke saath** (odd $n$) group karo:
$$e^{i\theta}=\underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right)}_{\text{real part}} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\text{imag part}}$$
*Yeh step kyun?* Rearrangement allowed hai kyunki series har $\theta$ ke liye **absolutely convergent** hai.
### Step 4 — Dono bracketed series ko pehchaano
Pehla bracket exactly $\cos\theta$ ki series hai; doosra exactly $\sin\theta$ hai (Step 0 se). Isliye
$$\boxed{\,e^{i\theta} = \cos\theta + i\sin\theta\,}$$
*Yeh step kyun?* Alternating even-power series **hi** cosine hai aur alternating odd-power series **hi** sine hai. Unhe term-for-term match karna proof complete karta hai. $\blacksquare$
> [!formula] Jo consequences free mein milte hain
> - $|e^{i\theta}| = \sqrt{\cos^2\theta+\sin^2\theta}=1$ (unit circle pe point).
> - $e^{-i\theta} = \cos\theta - i\sin\theta$ ($-\theta$ daalo).
> - $\cos\theta = \dfrac{e^{i\theta}+e^{-i\theta}}{2},\qquad \sin\theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}.$
> - **Euler's identity:** $\theta=\pi\Rightarrow e^{i\pi}=-1\Rightarrow e^{i\pi}+1=0.$
> - **De Moivre:** $(e^{i\theta})^n=e^{in\theta}\Rightarrow(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta.$
![[3.5.06-Euler's-formula-—-e^(iθ)-=-cos-θ-+-i-sin-θ-(proof-via-Taylor-series).png]]
---
## Worked examples
> [!example] 1 — $e^{i\pi/2}$ evaluate karo
> **Plan:** $\theta=\pi/2$ daalo.
> $$e^{i\pi/2}=\cos\tfrac{\pi}{2}+i\sin\tfrac{\pi}{2}=0+i(1)=i.$$
> *Yeh step kyun?* $\pi/2$ rad ek quarter-turn hai, jo unit circle ke top pe land karta hai, jo $i$ hai. ✔ "Multiplying by $i$ = 90° rotation" ke saath consistent.
> [!example] 2 — Prove karo $\cos(A+B)=\cos A\cos B-\sin A\sin B$
> **Plan:** $e^{i(A+B)}$ ko do tarike se compare karo.
> $$e^{i(A+B)}=e^{iA}e^{iB}=(\cos A+i\sin A)(\cos B+i\sin B).$$
> Expand karo: $=(\cos A\cos B-\sin A\sin B)+i(\sin A\cos B+\cos A\sin B).$
> Lekin $e^{i(A+B)}=\cos(A+B)+i\sin(A+B)$ bhi hai.
> **Real parts match karo:** $\cos(A+B)=\cos A\cos B-\sin A\sin B$. ✔ (imaginary parts se $\sin$ identity milti hai)
> *Yeh step kyun?* Ek hi complex number ke do expressions ke real aur imaginary parts barabar hone chahiye.
> [!example] 3 — $(1+i)^{8}$ compute karo
> **Plan:** $1+i$ ko polar form $re^{i\theta}$ mein likho, phir De Moivre use karo.
> $r=\sqrt{1^2+1^2}=\sqrt2$, $\theta=\pi/4$ (first quadrant, equal parts). Toh $1+i=\sqrt2\,e^{i\pi/4}$.
> $$(1+i)^8=(\sqrt2)^8 e^{i\cdot 8\pi/4}=16\,e^{i2\pi}=16(\cos2\pi+i\sin2\pi)=16.$$
> *Yeh step kyun?* Power mein uthana = modulus ko power mein uthao, angle ko **multiply** karo. $8\times\pi/4=2\pi$ = full turn, real axis par wapas.
---
> [!mistake] Steel-manned traps
> **Trap A: "$e^{i\theta}$ real $e^{x}$ ki tarah blow up karta hai."** *Kyun sahi lagta hai:* real exponentials grow karte hain. *Fix:* exponent **imaginary** hai, isliye stretch karne ki jagah, yeh rotate karta hai. $|e^{i\theta}|=1$ hamesha — yeh circle pe hamesha rehta hai.
>
> **Trap B: Degrees use karna.** Right angle ke liye $e^{i\cdot 90}$ likhna. *Kyun sahi lagta hai:* hum usually "90°" kehte hain. *Fix:* $\sin,\cos$ ki Taylor series tabhi in functions ke barabar hoti hain jab argument **radians** mein ho. $\theta=\pi/2$ use karo, $90$ nahi.
>
> **Trap C: "$\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2}$."** *Kyun sahi lagta hai:* cosine formula ke saath symmetry. *Fix:* subtract karne se real parts cancel ho jaate hain aur $2i\sin\theta$ bach jaata hai, isliye $2$ se nahi, $2i$ se divide karna padega.
>
> **Trap D: $(\cos\theta+i\sin\theta)^n=\cos\theta^n + \dots$** *Kyun sahi lagta hai:* lagta hai power distribute ho rahi hai. *Fix:* De Moivre **angle** ko multiply karta hai: $\cos n\theta + i\sin n\theta$.
---
> [!recall]- Feynman style: ek 12-saal ke bachche ko samjhao
> Ek ghadi ki sui imagine karo jiska length 1 hai aur jo centre pe pini hui hai. Number $\theta$ yeh hai ki *tum sui ko kitna ghumate ho* (radians mein). Jahan bhi tip land kare, uski zameen par shadow $\cos\theta$ hai (left–right) aur uski height $\sin\theta$ hai (up–down). Magical symbol $e^{i\theta}$ bas ek compact naam hai "sui ki tip $\theta$ se ghoomne ke baad." Ek ke baad ek do angles se ghoomna = angles add karna, aur isliye $e^{i A}\cdot e^{iB}=e^{i(A+B)}$: do spins ek saath stack ho jaate hain.
> [!mnemonic] Yaad rakho
> **"COSy real, SINful imaginary."** **Real** part **cos** hai, woh part jo $i$ se multiply hota hai (woh "imaginary/sinful" wala) **sin** hai. Aur $i$ ki sign story: **1, i, −1, −i** → "positive, up, negative, down" — chakkar lagata hua.
---
## #flashcards/maths
State Euler's formula. ::: $e^{i\theta}=\cos\theta+i\sin\theta$ (with $\theta$ in radians).
Euler's formula prove karne mein kaun si teen Taylor series combine hoti hain? ::: $e^x$, $\cos x$, $\sin x$ (Maclaurin series).
Proof mein terms real aur imaginary parts mein kyun split hoti hain? ::: Kyunki $i^n$ cycle karta hai $1,i,-1,-i$: even powers real terms dete hain, odd powers $i\times$real terms dete hain.
$|e^{i\theta}|$ kya hai? ::: Sabhi real $\theta$ ke liye exactly $1$ (yeh unit circle pe rehta hai).
$\cos\theta$ aur $\sin\theta$ ko exponential form mein likho. ::: $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$, $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$.
Euler's identity state karo. ::: $e^{i\pi}+1=0$.
$e^{i\pi/2}$ compute karo. ::: $i$.
Euler's formula se De Moivre's theorem state karo. ::: $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$, kyunki $(e^{i\theta})^n=e^{in\theta}$.
$\theta$ radians mein kyun hona chahiye? ::: $\sin,\cos$ ki power series un functions ke barabar tabhi hoti hain jab argument radian mein ho.
Exponentials mein $\sin\theta$ ka common galat version aur fix? ::: Galat: $2$ se divide karo. Fix: $2i$ se divide karo, kyunki subtraction se $2i\sin\theta$ bachta hai.
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## Connections
- [[Complex Numbers — Polar / Modulus-Argument form]]
- [[De Moivre's Theorem]]
- [[Roots of Unity]]
- [[Taylor & Maclaurin Series]]
- [[Multiplication as Rotation & Scaling]]
- [[Euler's Identity $e^{i\pi}+1=0$]]
- [[Hyperbolic Functions — $\cosh,\sinh$ vs $\cos,\sin$]]
## 🖼️ Concept Map
```mermaid
flowchart TD
ROT[Multiply by i is 90 degree rotation]
EULER[Euler's formula e^itheta = cos + i sin]
SERIES[Maclaurin series of e^x cos x sin x]
ENTIRE[Functions are entire, converge for all complex]
SUB[Substitute x = i theta]
CYCLE[Powers of i cycle with period 4]
SPLIT[Split into real and imaginary parts]
COS[Real part equals cos theta]
SIN[Imag part equals sin theta]
CIRCLE[Point on unit circle at angle theta]
DEMOIVRE[De Moivre's theorem]
IDENTITY[e^i pi + 1 = 0]
ROT -->|intuition for| EULER
SERIES -->|defined for complex via| ENTIRE
ENTIRE -->|justifies| SUB
SUB -->|applied to e^x| CYCLE
CYCLE -->|enables| SPLIT
SPLIT -->|gives| COS
SPLIT -->|gives| SIN
COS -->|combine to| EULER
SIN -->|combine to| EULER
EULER -->|describes| CIRCLE
EULER -->|yields| DEMOIVRE
EULER -->|special case| IDENTITY
```
## 🔬 Deep Dive
> [!intuition] Aur gehraai mein jao — visual, zero se
> Is topic ki step-by-step 3Blue1Brown-style breakdowns.
- [[3.5.06 D1 Foundations|D1 · Foundations — har symbol zero se]]
- [[3.5.06 D2 Visual Walkthrough|D2 · Visual walkthrough — derivation pictures mein]]
- [[3.5.06 D3 Worked Examples|D3 · Worked examples — har scenario]]
- [[3.5.06 D4 Exercises|D4 · Exercises — graded, full solutions]]
- [[3.5.06 D5 Question Bank|D5 · Question bank — concept traps]]