Visual walkthrough — Polar form — r(cos θ + i sin θ) = r·cis θ

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Step 1 — Plant the arrow

WHY. Numbers are hard to "spin" and "stretch"; arrows are easy to picture spinning and stretching. Turning $z$ into an arrow lets geometry do the work that algebra found awkward.

In the picture, term by term:

• $O$ = the origin $(0,0)$, the tail of every arrow.
• $a$ = how far right the tip sits (the red horizontal distance).
• $b$ = how far up the tip sits (the mint vertical distance).
• The lavender arrow is $z$ itself.

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Step 2 — Manufacture a right triangle

WHY this exact move? Trigonometry — the machinery that links angles to side ratios — is only defined on right triangles. We have an angle $\theta$ we want to understand, so we deliberately build the one shape where angle-tools work.

Reading the triangle:

• Bottom leg (horizontal, coral) has length $a$ — it lies flat along the real axis.
• Right leg (vertical, mint) has length $b$ — it stands straight up.
• Hypotenuse (lavender) is the arrow, length $r$.
• The angle between the bottom leg and the hypotenuse is our $\theta$.
• The corner where the vertical meets the horizontal is a perfect $90^{\circ }$ (the little square).

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Step 3 — Read the trig ratios off the triangle

WHY these two tools and not $\tan$ here? We want to eventually recover both $a$ and $b$ separately. $\cos$ reaches the horizontal leg $a$; $\sin$ reaches the vertical leg $b$. $\tan =\sin /\cos =b/a$ mixes them together and loses the length $r$ — great for finding the angle later, useless for rebuilding the arrow now. So here we pick the two ratios that each isolate one leg.

Term by term in $\cos \theta =\frac{a}{r}$:

• adjacent = the leg touching the angle $\theta$ that is not the hypotenuse → the bottom leg $a$.
• hypotenuse = the longest side, always opposite the right angle → the arrow $r$.
• So $\cos \theta$ measures "what fraction of the full length points sideways."

And in $\sin \theta =\frac{b}{r}$:

• opposite = the leg across from $\theta$ → the vertical leg $b$.
• So $\sin \theta$ measures "what fraction of the full length points upward."

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Step 4 — Solve for the coordinates

$a=r\cos \theta ,b=r\sin \theta$

WHY. Our goal is to replace the description $(a,b)$ with the description $(r,\theta )$. To do that we must express $a$ and $b$ in terms of $r$ and $\theta$. Multiplying each ratio through by $r$ does exactly that.

Term by term:

• In $a=r\cos \theta$: take the full length $r$, keep only its sideways fraction $\cos \theta$ → you land on the horizontal position $a$. ("$a$ rides cos.")
• In $b=r\sin \theta$: take the full length $r$, keep only its upward fraction $\sin \theta$ → you land on the vertical position $b$. ("$b$ rides sin.")

The figure shows the arrow of length $r$ casting a shadow of length $r\cos \theta$ on the floor and a shadow of length $r\sin \theta$ on the wall — the two projections that are $a$ and $b$.

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Step 5 — Substitute back and collapse to polar form

$z=a+bi=(r\cos \theta )+(r\sin \theta )i$

Now factor out the common $r$:

$z=r(\cos \theta +i\sin \theta )=r\text{cis}\theta ■$

WHY factor? Pulling $r$ (the size) outside separates it cleanly from $\cos \theta +i\sin \theta$ (the direction). The bracket is a pure "pointing" object of length exactly $1$; the $r$ in front just scales it.

Term by term in $r(\cos \theta +i\sin \theta )$:

• $r$ = the arrow's length — stretches the unit direction to the right size.
• $\cos \theta$ = the real part of the pure direction (its sideways component).
• $i\sin \theta$ = the imaginary part of the pure direction (its upward component), carried by $i$.
• $\text{cis}\theta$ = a nickname for the whole bracket: cos i sin. It is a sum, not a product.

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Step 6 — The degenerate cases (never skipped)

WHY we must do this. Every claim in Steps 2–5 was read off the triangle. But a triangle needs three genuine sides. If $b=0$ the vertical leg disappears; if $a=0$ the horizontal leg disappears; if $r=0$ there is no arrow at all. In each case the drawing collapses and the ratio $b/a$ can even become $0/0$ — so the general recipe is silent and we must check these boundary points directly, or a reader landing on the real axis or the origin would have no answer. Covering them is not optional decoration; it is what makes the formula total.

Case A — on the positive real axis ($b=0,a>0$). The arrow lies flat pointing right. No vertical leg. Here $\theta =0$, and indeed $r(\cos 0+i\sin 0)=r(1+0)=r=a$. ✓

Case B — on the negative real axis ($b=0,a<0$). Arrow points left. $\theta =\pi$, and $r(\cos \pi +i\sin \pi )=r(-1+0)=-r=a$. ✓

Case C — on the positive imaginary axis ($a=0,b>0$). Arrow points straight up. No horizontal leg. $\theta =\frac{\pi }{2}$, and $r(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})=r(0+i)=bi$. ✓

Case D — on the negative imaginary axis ($a=0,b<0$). Arrow points straight down. $\theta =-\frac{\pi }{2}$, and $r(0-i)=-ri=bi$. ✓

Case E — the origin ($z=0$). Length $r=0$, so the arrow is a single point with no direction. The argument $\theta$ is undefined — you cannot ask "which way does a zero-length arrow point?" This is the one complex number with no polar form.

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Step 7 — Quadrants: where the arrow really points

WHY this is a trap. $\tan$ takes the same value for an arrow and for that same arrow spun by $180^{\circ }$ (both legs flip sign, the ratio $b/a$ is unchanged). So ${\tan }^{-1}$ genuinely cannot tell "up-left" from "down-right." We repair it by reading the signs of $a$ and $b$ to see which corner — which quadrant — the point sits in.

Let $\alpha ={\tan }^{-1}\left|\frac{b}{a}\right|$ be the reference angle — a positive acute angle (between $0$ and $\frac{\pi }{2}$) measuring the arrow's steepness against the horizontal axis, ignoring direction.

Quadrant	Signs $(a,b)$	Where it points	Argument $\theta$
I	$(+,+)$	up-right	$\theta =\alpha$
II	$(-,+)$	up-left	$\theta =\pi -\alpha$
III	$(-,-)$	down-left	$\theta =-(\pi -\alpha )$
IV	$(+,-)$	down-right	$\theta =-\alpha$

WHY each fix. The reference angle $\alpha$ only knows steepness, not which corner. In Quadrant II the arrow leans up-left, so we take $\pi$ (pointing fully left) and back off by $\alpha$ → $\pi -\alpha$. In Quadrant III we mirror that below the axis → $-(\pi -\alpha )$. In Quadrant IV the arrow dips below the right axis → simply $-\alpha$. Each result already lands inside $(-\pi ,\pi ]$, so it is the principal argument.

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The one-picture summary

This final figure compresses the entire walkthrough: the arrow $z$ from the origin $O$, its triangle with legs $a=r\cos \theta$ and $b=r\sin \theta$, the length $r=\sqrt{a^2+b^2}$, the angle $\theta$, and the finished formula $z=r\text{cis}\theta$ — all in one frame. Every earlier step is a zoom into one part of this diagram.

Where this leads next: multiplying arrows becomes "multiply lengths, add angles" (Multiplication and Division of Complex Numbers), raising to powers becomes De Moivre's Theorem, and the deeper identity $e^{i\theta }=\cos \theta +i\sin \theta$ lives in Euler's Formula — all built directly on the picture above. Finding $n$-th roots (Roots of Complex Numbers) is the same picture run backwards.

Recall Feynman: tell the whole walkthrough to a 12-year-old

Stand in a field, at the spot we'll call the origin $O$. A treasure sits somewhere. First I describe it your way: "3 steps east, 4 steps north" — that's $a$ and $b$. Now I draw an arrow straight from me to the treasure and drop a line down to the ground. That makes a right triangle: the flat bottom is my east-distance, the vertical bit is my north-distance, and the slanted arrow is the actual straight-line distance $r$ — which by Pythagoras is $\sqrt{a^2+b^2}$. Trig is a machine that, if I know the arrow's length and its tilt angle $\theta$, tells me how much of it points east ($r\cos \theta$) and how much points north ($r\sin \theta$). So instead of "3 east, 4 north," I can say "5 steps, tilted at this angle." Same treasure. The only tricky bit: if the treasure is behind me or to my left, a plain calculator gets confused because it can't tell "front-left" from "back-right" — they tilt the same. So I first look at which corner the treasure is in, then adjust the angle. And if the treasure is right under my feet ($z=0$), there's no arrow at all — no direction to name.

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Connections

• Argand Diagram — the plane every figure here is drawn on.
• Modulus and Argument — the length $r$ and angle $\theta$ we built.
• Multiplication and Division of Complex Numbers — the payoff: add angles, multiply lengths.
• De Moivre's Theorem — powers of $\text{cis}\theta$.
• Euler's Formula — why $\text{cis}\theta$ behaves like $e^{i\theta }$.
• Roots of Complex Numbers — the derivation, run in reverse.