Euler's formula — e^(iθ) = cos θ + i sin θ (proof via Taylor series)

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WHAT is being claimed

WHY care? It converts multiplication of complex numbers into addition of angles, turns trig identities into one-line algebra, gives De Moivre's theorem instantly, and produces the most famous equation in maths: $e^{i\pi}+1=0$.

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HOW to derive it — from Taylor series (first principles)

WHY this method? We define $e^x$, $\cos x$, $\sin x$ by their power series (they converge for all complex inputs). Then we just plug in $x=i\theta$ and let the powers of $i$ do the sorting.

Step 0 — The three series we start from

$e^{x} = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+\cdots$ $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+\cdots$ $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots$

Why this step? These are the Maclaurin (Taylor at $0$) expansions. They are valid for complex arguments because the exponential/trig functions are entire (their series converge everywhere), so substituting $i\theta$ is legal.

Step 1 — The engine: powers of $i$ cycle

$i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,\ \dots$ so $i^n$ repeats with period $4$: $\;1,\ i,\ -1,\ -i,\ 1,\dots$

Why this step? This four-fold cycle is what will split the exponential series into a real part and an imaginary part.

Step 2 — Substitute $x = i\theta$ into $e^x$

= \sum_{n=0}^{\infty}\frac{i^n\,\theta^n}{n!}.$$ Write out terms, using $i^0{=}1,\ i^1{=}i,\ i^2{=}{-}1,\ i^3{=}{-}i,\dots$: $$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$$ *Why this step?* Each $i^n$ turns into $+1,+i,-1,-i,\dots$, so half the terms carry a factor of $i$ and half don't. ### Step 3 — Separate real ($i^{even}$) and imaginary ($i^{odd}$) terms Group the terms **without** $i$ (even $n$) and **with** $i$ (odd $n$): $$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}}$$ *Why this step?* Rearrangement is allowed because the series is **absolutely convergent** for every $\theta$. ### Step 4 — Recognise the two bracketed series The first bracket is exactly the series for $\cos\theta$; the second is exactly $\sin\theta$ (from Step 0). Therefore $$\boxed{\,e^{i\theta} = \cos\theta + i\sin\theta\,}$$ *Why this step?* The alternating even-power series **is** cosine and the alternating odd-power series **is** sine. Matching them term-for-term completes the proof. $\blacksquare$ > [!formula] Consequences you get for free > - $|e^{i\theta}| = \sqrt{\cos^2\theta+\sin^2\theta}=1$ (point on unit circle). > - $e^{-i\theta} = \cos\theta - i\sin\theta$ (put $-\theta$). > - $\cos\theta = \dfrac{e^{i\theta}+e^{-i\theta}}{2},\qquad \sin\theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}.$ > - **Euler's identity:** $\theta=\pi\Rightarrow e^{i\pi}=-1\Rightarrow e^{i\pi}+1=0.$ > - **De Moivre:** $(e^{i\theta})^n=e^{in\theta}\Rightarrow(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta.$ ![[3.5.06-Euler's-formula-—-e^(iθ)-=-cos-θ-+-i-sin-θ-(proof-via-Taylor-series).png]] --- ## Worked examples > [!example] 1 — Evaluate $e^{i\pi/2}$ > **Plan:** put $\theta=\pi/2$. > $$e^{i\pi/2}=\cos\tfrac{\pi}{2}+i\sin\tfrac{\pi}{2}=0+i(1)=i.$$ > *Why this step?* $\pi/2$ rad is a quarter-turn, landing on the top of the unit circle, which is $i$. ✔ Consistent with "multiplying by $i$ = 90° rotation." > [!example] 2 — Prove $\cos(A+B)=\cos A\cos B-\sin A\sin B$ > **Plan:** compare $e^{i(A+B)}$ two ways. > $$e^{i(A+B)}=e^{iA}e^{iB}=(\cos A+i\sin A)(\cos B+i\sin B).$$ > Expand: $=(\cos A\cos B-\sin A\sin B)+i(\sin A\cos B+\cos A\sin B).$ > But also $e^{i(A+B)}=\cos(A+B)+i\sin(A+B)$. > **Match real parts:** $\cos(A+B)=\cos A\cos B-\sin A\sin B$. ✔ (imag parts give the $\sin$ identity) > *Why this step?* Two expressions for the same complex number must have equal real and equal imaginary parts. > [!example] 3 — Compute $(1+i)^{8}$ > **Plan:** write $1+i$ in polar form $re^{i\theta}$, then use De Moivre. > $r=\sqrt{1^2+1^2}=\sqrt2$, $\theta=\pi/4$ (first quadrant, equal parts). So $1+i=\sqrt2\,e^{i\pi/4}$. > $$(1+i)^8=(\sqrt2)^8 e^{i\cdot 8\pi/4}=16\,e^{i2\pi}=16(\cos2\pi+i\sin2\pi)=16.$$ > *Why this step?* Raising to a power = raise modulus to the power, **multiply** the angle. $8\times\pi/4=2\pi$ = full turn back to the real axis. --- > [!mistake] Steel-manned traps > **Trap A: "$e^{i\theta}$ blows up like $e^{x}$ does."** *Why it feels right:* real exponentials grow. *Fix:* the exponent is **imaginary**, so instead of stretching, it rotates. $|e^{i\theta}|=1$ always — it stays on the circle forever. > > **Trap B: Using degrees.** Writing $e^{i\cdot 90}$ for a right angle. *Why it feels right:* we usually say "90°". *Fix:* the Taylor series of $\sin,\cos$ only equal these functions when the argument is in **radians**. Use $\theta=\pi/2$, not $90$. > > **Trap C: "$\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2}$."** *Why it feels right:* symmetry with the cosine formula. *Fix:* subtracting cancels the real parts and leaves $2i\sin\theta$, so you must divide by $2i$, not $2$. > > **Trap D: $(\cos\theta+i\sin\theta)^n=\cos\theta^n + \dots$** *Why it feels right:* looks like distributing a power. *Fix:* De Moivre multiplies the **angle**: $\cos n\theta + i\sin n\theta$. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a clock hand of length 1 pinned at the centre. The number $\theta$ is *how far you turn the hand* (in radians). Wherever the tip lands, its shadow on the floor is $\cos\theta$ (left–right) and its height is $\sin\theta$ (up–down). The magic symbol $e^{i\theta}$ is just a compact name for "the tip of the hand after turning by $\theta$." Turning by two angles one after another = adding the angles, and that's why $e^{i A}\cdot e^{iB}=e^{i(A+B)}$: two spins stack up. > [!mnemonic] Remember it > **"COSy real, SINful imaginary."** The **real** part is **cos**, the part multiplied by $i$ (the "imaginary/sinful" bit) is **sin**. And the sign story of $i$: **1, i, −1, −i** → "positive, up, negative, down" spinning around. --- ## #flashcards/maths State Euler's formula. ::: $e^{i\theta}=\cos\theta+i\sin\theta$ (with $\theta$ in radians). Which three Taylor series are combined to prove Euler's formula? ::: $e^x$, $\cos x$, $\sin x$ (Maclaurin series). Why do the terms split into real and imaginary parts in the proof? ::: Because $i^n$ cycles $1,i,-1,-i$: even powers give real terms, odd powers give $i\times$real terms. What is $|e^{i\theta}|$? ::: Exactly $1$ for all real $\theta$ (it lies on the unit circle). Write $\cos\theta$ and $\sin\theta$ in exponential form. ::: $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$, $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$. State Euler's identity. ::: $e^{i\pi}+1=0$. Compute $e^{i\pi/2}$. ::: $i$. State De Moivre's theorem from Euler's formula. ::: $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$, since $(e^{i\theta})^n=e^{in\theta}$. Why must $\theta$ be in radians? ::: The power series for $\sin,\cos$ equal those functions only for radian arguments. Common wrong version of $\sin\theta$ in exponentials and the fix? ::: Wrong: divide by $2$. Fix: divide by $2i$, because subtraction leaves $2i\sin\theta$. --- ## 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 ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, Euler's formula ka core idea bahut simple hai: $e^{i\theta}$ ka matlab hai unit circle par $\theta$ radian ghoom jaana. Agar aap ek 1 length ki suyi (clock hand) ko $\theta$ angle se rotate karo, to uska tip jahan pahunchta hai uska horizontal component $\cos\theta$ hota hai aur vertical component $\sin\theta$. Isliye $e^{i\theta}=\cos\theta+i\sin\theta$. Yaad rakho — real part cosine, aur $i$ wala part sine. > > Proof kaise aata hai? Hum teen Taylor series lete hain: $e^x$, $\cos x$, aur $\sin x$. Ab $x$ ki jagah $i\theta$ daal do. $i$ ke powers ek cycle mein chalte hain: $1, i, -1, -i$, phir wapas $1$. Isi cycle ki wajah se $e^{i\theta}$ ki series do hisson mein tut jaati hai — jo terms bina $i$ ke hain wo cosine ki series ban jaati hain, aur jo $i$ ke saath hain wo sine ki series. Bas, dono ko match kar do, formula proved! > > Ye formula kyun important hai? Kyunki complex numbers ko multiply karna angle add karne jaisa ban jaata hai. Jaise $(1+i)^8$ jaisa bada calculation, polar form $\sqrt2\,e^{i\pi/4}$ mein likhkar seconds mein $16$ nikal jaata hai — bina jhanjhat. De Moivre theorem, roots of unity, aur trig identities sab isi ek line se aa jaate hain. > > Ek warning: hamesha radians use karo, degrees nahi — kyunki series sirf radians mein sahi hoti hai. Aur $|e^{i\theta}|$ hamesha $1$ hota hai, ye kabhi $e^x$ ki tarah bada nahi hota, kyunki imaginary exponent stretch nahi rotate karta hai. Isko dil se samajh lo, to complex numbers ka aadha chapter apne aap clear ho jayega. ![[audio/3.5.06-Euler's-formula-—-e^(iθ)-=-cos-θ-+-i-sin-θ-(proof-via-Taylor-series).mp3]]