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

Before we start, three plain-word promises about the cast of characters:

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Step 1 — Draw the plane, and see what $i$ is

WHAT. We lay out the two directions: rightward (real) and upward (imaginary). The number $1$ sits one unit right of the centre. The number $i$ sits one unit up.

WHY. Everything that follows is a journey of one point across this sheet. We must first agree on the map: which way is real, which way is imaginary, and where $0$, $1$, $i$ live.

PICTURE. Look at the figure: the blue axis is the real number line you already know. The orange axis is brand new — it is the "$\times i$" direction. The green dot at $(0,1)$ is $i$ itself.

$\underset{\text{steps right}}{\underbrace{x}}+\underset{{\text{turn 90}}^{\circ }}{\underbrace{i}}\underset{\text{steps up}}{\underbrace{y}}$

• $x$ — how far right, along the blue axis.
• $y$ — how far up, along the orange axis.
• $i$ — the tag that says "this part points up, not right".

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Step 2 — Multiplying by $i$ is a quarter-turn

WHAT. Take any point and multiply it by $i$. We claim the point rotates 90° counter-clockwise about the centre, keeping its distance from the centre unchanged.

WHY. This single fact is the seed of the whole formula. If growing "in the $i$ direction" means turning, then repeatedly nudging in that direction should make us go around — a circle. We need to see that $\times i$ = turn.

PICTURE. Start at $1$ (right). Multiply by $i$: land on $i$ (up) — a quarter turn. Multiply again: $i\cdot i=i^2=-1$ (left) — another quarter turn. Again: $-i$ (down). Again: back to $1$. The four coloured arrows in the figure spin around a square path.

$1\stackrel{\times i}{\to }i\stackrel{\times i}{\to }-1\stackrel{\times i}{\to }-i\stackrel{\times i}{\to }1$

• Each arrow is one multiplication by $i$.
• The length never changes (still distance $1$ from centre) — only the direction turns by $90^{\circ }$.
• After four turns we are home: this is exactly the cycle $i^0,i^1,i^2,i^3=1,i,-1,-i$ that powers the algebraic proof.

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Step 3 — What does $e^{\theta }$ mean, before we make it imaginary?

WHAT. The symbol $e^t$ is the answer to "start at $1$ and grow at a rate equal to your current size". A key fact from calculus: the velocity of $e^t$ equals its position.

WHY. We are about to put an $i$ into the exponent. To understand what that does, we first need to know what the exponent controls: it controls the direction and size of the push at each instant. For plain $e^t$ the push points the same way you already are, so you shoot off to the right and grow forever.

PICTURE. The blue dot moves right; the red push-arrow points the same way as its position, so it accelerates away. Nothing curves — the push and the position are parallel.

$\frac{d}{dt}{e}^t=e^t$

• Left side — the push (velocity) of the moving point.
• Right side — the point's current position.
• They are equal ⟹ push is always along the position ⟹ pure growth, no turning.

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Step 4 — Put an $i$ in front: the push turns sideways

WHAT. Now consider $e^{it}$: a point whose position starts at $1$ and whose push is $i$ times its position.

WHY. By Step 2, multiplying by $i$ rotates by $90^{\circ }$. So the push is no longer along the position — it is perpendicular to it. A perpendicular push cannot make you longer; it can only steer you sideways. Sideways-forever means circling.

PICTURE. The blue arrow is the position (from centre to dot). The orange arrow is the push $i\cdot (\text{position})$ — rotated $90^{\circ }$ from blue, tangent to the circle. Because it is exactly perpendicular, the dot slides along a circle of radius $1$ and never leaves it.

$\frac{d}{dt}{e}^{it}=i{e}^{it}$

• $e^{it}$ — current position (blue).
• $i{e}^{it}$ — the push (orange), that same vector turned $90^{\circ }$.
• Perpendicular push ⟹ constant length ⟹ motion on the unit circle at speed $1$.

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Step 5 — Read off the coordinates: cos and sin appear

WHAT. After sweeping angle $\theta$ around the unit circle starting from $1$, where is the dot? Its horizontal coordinate is $\cos \theta$ and its vertical coordinate is $\sin \theta$ — that is the very definition of cosine and sine.

WHY. We've now placed the point $e^{i\theta }$ on the circle at angle $\theta$. To write it as a complex number $x+iy$ we just read its shadow on each axis. Those shadows are what "cosine" and "sine" have always meant.

PICTURE. Drop a vertical line from the dot to the real axis: that foot is at $\cos \theta$ (blue segment). The height of the dot is $\sin \theta$ (orange segment). The right triangle has hypotenuse $1$.

$e^{i\theta }=\underset{\text{shadow on real axis}}{\underbrace{\cos \theta }}+i\underset{\text{height, up-direction}}{\underbrace{\sin \theta }}$

• $\cos \theta$ — how far right the dot sits (adjacent side ÷ hypotenuse $1$).
• $\sin \theta$ — how far up the dot sits (opposite side ÷ hypotenuse $1$).
• The $i$ tags the height as the imaginary part. This is Euler's formula. $■$

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Step 6 — Match the pictures to the Taylor-series proof

WHAT. The parent note proves the same thing by expanding $e^{i\theta }=\sum \frac{(i\theta {)}^n}{n!}$ and splitting even/odd powers. Here is how that algebra is the spinning motion.

WHY. Two proofs, one truth — seeing they agree cements it. The four-step cycle $1,i,-1,-i$ from the series is literally the four quarter-turns of Step 2.

PICTURE. Each term $\frac{(i\theta {)}^n}{n!}$ is a small arrow. The first arrow points right ($1$), the next up ($i\theta$), the next left, the next down — each rotated $90^{\circ }$ from the last, each shorter (divided by a bigger factorial). Chain them tip-to-tail and the broken path curls onto the circle, landing at the dot from Step 5.

$e^{i\theta }=\underrightarrow{\underbrace{1}}+\underset{\uparrow }{\underbrace{i\theta }}+\underleftarrow{\underbrace{\frac{(i\theta {)}^2}{2!}}}+\underset{\downarrow }{\underbrace{\frac{(i\theta {)}^3}{3!}}}+\cdots$

• Each term inherits its direction from $i^n$: right, up, left, down, repeating — the same cycle as the quarter-turns.
• Grouping the right/left terms gives $\cos \theta$; grouping the up/down terms gives $\sin \theta$. Same destination, drawn as a staircase of shrinking arrows.

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Step 7 — The edge cases: check every landmark

WHAT. A formula is only trustworthy if it survives the special inputs. We test $\theta =0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi$, and negative $\theta$.

WHY. These are the quadrant boundaries. If the moving dot lands on the wrong spot at even one of them, the formula is broken. The reader should never meet an angle we didn't show.

PICTURE. Five labelled dots around the circle, plus a dashed clockwise arc for negative angles.

$\theta$	position on circle	$e^{i\theta }=\cos \theta +i\sin \theta$
$0$	right	$1+0i=1$
$\frac{\pi }{2}$	top	$0+i(1)=i$
$\pi$	left	$-1+0i=-1$
$\frac{3\pi }{2}$	bottom	$0+i(-1)=-i$
$2\pi$	right again	$1$ (full loop)

• $\theta =0$ (degenerate case): no turn at all — the dot stays at $1$. The formula gives $1$. ✔
• $\theta =\pi$: half a turn to the left, giving $-1$ — this is Euler's identity. ✔
• Negative $\theta$ (clockwise): the push $i{e}^{it}$ still turns the same way, but time runs backward, so we spin clockwise: $e^{-i\theta }=\cos \theta -i\sin \theta$ (height flips sign, shadow unchanged). ✔
• $\theta >2\pi$ or beyond one loop: we simply keep circling — the position repeats every $2\pi$, matching that $\cos ,\sin$ have period $2\pi$. ✔

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The one-picture summary

WHAT. One figure holds the whole story: the perpendicular push (Step 4) steering the position (Step 3) around the unit circle, casting shadows $\cos \theta$ and $\sin \theta$ (Step 5), with the tip labelled $e^{i\theta }$.

$\boxed{e^{i\theta }=\cos \theta +i\sin \theta }\text{position at angle }\theta ,\text{push}=i\times \text{position}.$

Recall Feynman retelling — the whole walkthrough in plain words

Picture a clock hand of length $1$, pinned at the middle of a flat sheet. "Multiply by $i$" is a magic instruction: turn a quarter circle to the left. Ordinary growth, $e^t$, pushes a dot straight ahead so it races off to the right and gets bigger. But the instant we glue an $i$ onto the exponent, the push turns sideways — it's always pointing $90^{\circ }$ off from where the dot is. A sideways push can never make the string longer; it can only swing the dot around. So the tip glides around a circle of radius $1$ at a steady speed. Wait a time $\theta$ and you've swept exactly $\theta$ radians. Look at where the tip landed: its distance to the right is called $\cos \theta$, its height is called $\sin \theta$. So "the tip after turning by $\theta$", written $e^{i\theta }$, equals $\cos \theta +i\sin \theta$. The Taylor-series proof draws the same trip as a staircase of tiny arrows — right, up, left, down, each shorter than the last — that curls onto the very same circle.

Recall Quick self-check

What kind of push produces circular motion instead of growth? ::: A push perpendicular to the position (multiplication by $i$ turns it $90^{\circ }$). Why does the radius stay exactly $1$? ::: The perpendicular push never changes the distance from the centre; we started at $1$. After sweeping angle $\theta$, what are the dot's coordinates? ::: $(\cos \theta ,\sin \theta )$, i.e. $e^{i\theta }=\cos \theta +i\sin \theta$. Where does $e^{-i\theta }$ land compared to $e^{i\theta }$? ::: Mirror image across the real axis (spins clockwise): $\cos \theta -i\sin \theta$.

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Connections

• Euler's Formula — $e^{i\theta} = \cos\theta + i\sin\theta$ (proof via Taylor series)
• 3.5.06 D1 Foundations
• 3.5.06 D3 Worked Examples
• Multiplication as Rotation & Scaling
• Complex Numbers — Polar / Modulus-Argument form
• Taylor & Maclaurin Series
• Euler's Identity $e^{i\pi}+1=0$
• De Moivre's Theorem