Visual walkthrough — Imaginary unit i = √(−1), i² = −1, powers of i cycle

Before we start, two words in plain language:

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Step 1 — Put the real numbers on a line, then discover there is nowhere for $i$

WHAT. We draw the number line and mark where "squaring" sends things.

WHY. The whole reason $i$ exists is that $x^2=-1$ has no answer among the numbers we already have. Before inventing anything, we must see that gap. If we can see there is genuinely no spot on the line whose square is $-1$, we've earned the right to build a new place for it.

PICTURE. In the figure, take any point $x$ on the horizontal line and follow the arrow to $x^2$. Whether $x$ is positive (blue) or negative (pink), the arrow always lands on the non-negative side (the pale-yellow zone $\ge 0$). The point $-1$ (left of $0$) is never a landing spot. That empty target is the hole $i$ will fill.

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Step 2 — Invent a second direction and place $i$ there

WHAT. We add a vertical ruler through $0$ and define the point one unit up to be $i$.

WHY. Since no spot on the horizontal line works, we make a new spot off it. Choosing "straight up, one unit" is not random: we want multiplying by $i$ to be a clean turn, and a quarter-turn from "east" (the point $1$) lands exactly on "north." We will justify that turn in Step 3; for now, just plant the flag.

PICTURE. The east arrow points to $1$ on the horizontal ruler. The north arrow points to $i$ on the new vertical ruler. West is $-1$, south is $-i$. Four compass points, all one unit from the centre.

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Step 3 — Show that multiplying by $i$ is a $90^{\circ }$ turn

WHAT. We check what "$\times i$" does to the starting point $1$, and then to $i$ itself.

WHY. Everything about the cycle falls out of one fact: multiplying by $i$ rotates a point $90^{\circ }$ anticlockwise. We need to earn that fact, not assume it. We test it on the point $1$ (multiplying $1$ by $i$ gives $i$ — one quarter-turn from east to north) and on $i$ (multiplying $i$ by $i$ gives $i^2=-1$ — another quarter-turn from north to west). Two turns match two multiplications, so the pattern is real.

PICTURE. The pink curved arrow shows $1\stackrel{\times i}{\to }i$: a $90^{\circ }$ swing from east to north. The blue curved arrow shows $i\stackrel{\times i}{\to }-1$: another $90^{\circ }$ swing from north to west. Same-size turn each time.

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Step 4 — Build the powers by multiplying step by step

WHAT. Starting from $i^0=1$, we multiply by $i$ over and over, each time replacing $i^2$ by $-1$ whenever it appears.

WHY. A "power" $i^n$ just means "multiply $n$ copies of $i$ together." Instead of memorising, we walk it: each new power is the previous one turned one more quarter-turn. Whenever two $i$'s collide into $i^2$, we swap in $-1$ — that is the only algebra move we ever make.

PICTURE. Follow the numbered chalk arrows around the circle: $i^0=1$ (east) $\to i^1=i$ (north) $\to i^2=-1$ (west) $\to i^3=-i$ (south) $\to i^4$ back to east.

$i^0=\underset{\text{no turns yet}}{\underbrace{1}}{i}^1=\underset{\text{one turn}}{\underbrace{i}}$ $i^2=i\cdot i=\underset{\text{by the rule}}{\underbrace{-1}}{i}^3=\underset{=-1}{\underbrace{i^2}}\cdot i=(-1)\cdot i=\underset{\text{three turns}}{\underbrace{-i}}$

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Step 5 — The fourth multiplication closes the loop

WHAT. We compute $i^4$ and find it equals $1$ — the starting point.

WHY. This is the heart of the whole topic. Four quarter-turns is $4\times 90^{\circ }=360^{\circ }$, a full circle, so you must arrive exactly where you began. Algebraically, $i^4=i^2\cdot i^2=(-1)(-1)=1$. Geometry and algebra agree: after four steps, reset.

PICTURE. The full pale-yellow circle is swept out; the dashed arc from south back to east completes the $360^{\circ }$. The label reads "$i^4=1$: home again."

$i^4=\underset{-1}{\underbrace{i^2}}\cdot \underset{-1}{\underbrace{i^2}}=(-1)(-1)=\underset{\text{back to start}}{\underbrace{1}},4\times 90^{\circ }=\underset{\text{full circle}}{\underbrace{360^{\circ }}}.$

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Step 6 — Any power reduces to one of four (the remainder rule)

WHAT. For a big exponent $n$, we split it into "full loops" plus a "leftover," and only the leftover matters.

WHY. Since every 4 steps brings you home, walking $n$ steps is the same as throwing away complete loops of 4 and walking only what remains. We write $n=4k+r$, where $k$ is how many complete loops and $r$ (one of $0,1,2,3$) is the leftover. Whole loops multiply the answer by $i^4=1$, which changes nothing — so the answer is just $i^r$.

PICTURE. A dial marked $0,1,2,3$: a big exponent like $27$ spins around several full times and settles on the leftover slot $3$, pointing at $-i$.

$i^n=i^{4k+r}=\underset{=1^k=1}{\underbrace{(i^4{)}^k}}\cdot \underset{\text{the leftover}}{\underbrace{i^r}}=i^r,r=n\mathrm{mod}4.$

• $\underset{\text{full loops}}{\underbrace{4k}}$ contributes $(i^4{)}^k=1$ — a picture of returning home $k$ times.
• $\underset{\text{remainder}}{\underbrace{r}}$ is the final short walk that decides where you stop.

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Step 7 — Running the loop backwards (negative powers)

WHAT. We ask what $i^{-1}$ (one divide by $i$) is, and see it turns the other way.

WHY. Dividing by $i$ should undo one multiplication — a $90^{\circ }$ turn clockwise instead of anticlockwise. We confirm this by clearing $i$ from the bottom of a fraction (the same trick as rationalising a surd): multiply top and bottom by $i$.

PICTURE. A green clockwise arrow: $1\stackrel{÷i}{\to }-i$ (east down to south), the mirror of Step 3's anticlockwise swing.

$i^{-1}=\frac{1}{i}=\frac{1}{i}\cdot \underset{=1,\text{clears the }i}{\underbrace{\frac{i}{i}}}=\frac{i}{\underset{-1}{\underbrace{i^2}}}=\frac{i}{-1}=\underset{\text{one turn clockwise}}{\underbrace{-i}}.$

So the loop also runs in reverse: $i^{-1}=-i,i^{-2}=-1,i^{-3}=i,i^{-4}=1$ — same four values, walked the other way.

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The one-picture summary

Everything on one board: the compass of four values, the anticlockwise arrows of "$\times i$," the clockwise arrows of "$÷i$," the $90^{\circ }$ labels, and the remainder dial that sends any exponent to one of four slots.

Recall Feynman retelling — the whole walkthrough in plain words

We started on the ordinary number line and noticed there's no spot whose square is $-1$: squaring always lands on the non-negative side. So we built a brand-new spot one unit straight up and named it $i$. Then we watched what "multiply by $i$" does: it swings a point a quarter-turn to the left. From the point $1$ (east), one swing lands on $i$ (north), the next on $-1$ (west), the next on $-i$ (south), and the fourth brings you all the way around — a full $360^{\circ }$ — right back to $1$. That is why the powers of $i$ only ever show four faces: $1,i,-1,-i$. For a giant exponent, you don't take every step; you throw away the complete four-loops (they each multiply by $1$ and change nothing) and only walk the leftover $0,1,2,$ or $3$ steps. Dividing by $i$ is the same dance in reverse — a quarter-turn clockwise. Four turns, one small circle, an entire chapter of algebra you can now see.

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Connections

• Complex Numbers — the framework $a+bi$ these turns live inside.
• Argand Diagram — the plane we drew from Step 2 onward.
• Modulus and Argument — each "$\times i$" adds $90^{\circ }$ to the argument.
• Roots of Unity — $\{1,i,-1,-i\}$ are the four 4th roots of unity.
• Euler's Formula — $i=e^{i\pi /2}$, so multiplying by $i$ adds $\pi /2$ to the angle.
• Solving Quadratic Equations — negative discriminants force us to use $i$.