Exponential form z = re^(iθ): Visual walkthrough

Step 0 — The three characters we will meet

Before any derivation, let me name the objects. We only need three ideas, each a picture.

Look at the picture: the black dot is the number $x + i y$ . Its horizontal shadow is $x$ (on the real axis); its vertical shadow is $y$ (on the imaginary axis). Everything else on this page is about where that dot goes.

Step 1 — Give $e^{u}$ a meaning that survives an imaginary input

WHAT. We replace the definition " $e$ times itself $u$ times" with an equivalent one that works for any input — a never-ending sum.

$e^{u} = start 1 + grow u + correction \frac{u ^{2}}{2 !} + \frac{u ^{3}}{3 !} + \frac{u ^{4}}{4 !} + \dots$

Each symbol: $u$ is whatever we feed in; $2! = 2 \times 1$ , $3! = 3 \times 2 \times 1$ (called a factorial). Notice the factorials grow explosively — $10! = 3, 628, 800$ — so from some point on, each new term is a tiny fraction of the one before it.

This exact sum is the Taylor and Maclaurin series of $e^{u}$ .

WHY this tool and not another? "Multiply $e$ by itself $i θ$ times" is nonsense — you can't repeat something an imaginary number of times. But a sum only needs adding and multiplying, which imaginary numbers can do. So we switch to the sum: it is the one description of $e^{u}$ that keeps working when $u$ turns imaginary.

PICTURE. For a real input each term is a positive push to the right, and the running total climbs toward $e^{u}$ .

The red staircase is the running total $1, 1 + u, 1 + u + \frac{u ^{2}}{2}, \dots$ landing on the true value (red dashed line) for $u = 1$ , where $e^{1} = e \approx 2.718$ . Notice: real input, real output, and it grows. Hold that thought — it changes completely in Step 3.

Step 2 — The one fact about $i$ that makes everything cycle

WHAT. We record what happens to $i$ when we keep multiplying it by itself.

$i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = + 1, i^{5} = i, \dots$

Each symbol: $i$ means "quarter-turn anticlockwise." Multiplying by $i$ again turns another quarter. Four quarter-turns is a full circle, back to the start — that is why $i^{4} = 1$ .

WHY we need this. In Step 3 we will feed $u = i θ$ into the sum from Step 1, so every term contains a power of $i$ . If those powers cycled, the whole sum will split into a repeating pattern. This step is the engine of that split.

PICTURE. Powers of $i$ march around the four compass points and repeat forever — the "cycle of 4."

Follow the red arrows: right $\to$ up $\to$ left $\to$ down $\to$ right. The pattern of signs $+, +, -, -$ (real vs imaginary) is exactly what will carve $cos$ and $sin$ out of one sum.

Step 3 — Feed $u = i θ$ in and let the powers of $i$ do their work

WHAT. Put $u = i θ$ into the Step 1 sum. Here $θ$ (theta) is a real number — an angle in radians (radians = arc-length on a circle of radius 1; a full turn is $2 π$ ).

$e^{i θ} = 1 + i θ + \frac{( i θ ) ^{2}}{2 !} + \frac{( i θ ) ^{3}}{3 !} + \frac{( i θ ) ^{4}}{4 !} + \dots$

Now replace each power of $i$ using Step 2 — $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ . Writing every $+$ and $-$ explicitly so the signs are unmistakable:

$e^{i θ} = real (+ 1) + imag (+ i θ) + real (- \frac{θ ^{2}}{2 !}) + imag (- i \frac{θ ^{3}}{3 !}) + real (+ \frac{θ ^{4}}{4 !}) + \dots$

Each bracket carries its own sign, and the label beneath says whether it lands on the real (horizontal) axis or the imaginary (vertical) axis — decided purely by the $i$ -cycle from Step 2.

WHY this step. This is the whole reason we rebuilt $e^{u}$ as a sum: an imaginary input is now perfectly legal, and the cycle of $i$ automatically sorts the terms into two families.

PICTURE. Alternating terms flip between the two axes — real, imaginary, real, imaginary — never blowing up, just nudging around.

Red terms sit on the imaginary axis, black terms on the real axis. Two clean piles are forming. Step 4 names them.

Step 4 — The two piles ARE cosine and sine

WHAT. Collect the real terms into one bracket and the imaginary terms into another:

$e^{i θ} = real pile (1 - \frac{θ ^{2}}{2 !} + \frac{θ ^{4}}{4 !} - \dots) + i imaginary pile (θ - \frac{θ ^{3}}{3 !} + \frac{θ ^{5}}{5 !} - \dots)$

WHY this is the finish. Those two brackets are literally the Taylor series of $cos θ$ and $sin θ$ (again from Taylor and Maclaurin series). $cos θ$ is the horizontal reach of a unit arrow tilted by $θ$ ; $sin θ$ is its vertical reach. So:

$e^{i θ} = cos θ + i sin θ$

Symbol by symbol: $cos θ$ is the real part (how far right), $sin θ$ is the coefficient of $i$ — the imaginary part (how far up).

PICTURE. The two piles are the horizontal and vertical legs of the arrow to a point on the unit circle.

The red arrow points to $e^{i θ}$ . Its horizontal leg is $cos θ$ , its vertical leg is $sin θ$ . This is the same right-triangle picture the parent's Polar form (trigonometric) r(cosθ + i sinθ) uses — we just derived it instead of assuming it.

Step 5 — Why it ROTATES and never GROWS (the length is always 1)

WHAT. Measure how far $e^{i θ}$ sits from the origin. Distance in the plane is $(right)^{2} + (up)^{2}$ :

$e^{i θ} = cos^{2} θ + sin^{2} θ = 1 = 1.$

The step $cos^{2} θ + sin^{2} θ = 1$ is Pythagoras on the Step 4 triangle: the two legs squared add to the hypotenuse squared, and the hypotenuse here is $1$ .

WHY this matters. For a real input (Step 2 picture) $e^{u}$ grew. For an imaginary input it does the opposite: it stays glued to the unit circle (the circle of radius $1$ ). Feeding in $i θ$ turned "growth" into "pure turning." This kills the classic mistake " $e^{i θ}$ must be huge."

PICTURE. As $θ$ increases, the red dot slides around the unit circle — same distance from centre forever.

The red arrow keeps length $1$ ; only its angle changes. That is the entire personality of $e^{i θ}$ : an angle dial, not a size dial.

Step 6 — Attach a size $r$ and get the full form $z = r e^{i θ}$

WHAT. Multiply the unit arrow $e^{i θ}$ by a positive number $r$ :

$z = r e^{i θ} = r cos θ + i r sin θ .$

Symbols: $r$ (the modulus, from Modulus and argument of a complex number) stretches the arrow's length to $r$ ; $θ$ (the argument) still sets only the direction. So $x = r cos θ$ and $y = r sin θ$ .

WHY. A general complex number needs two dials — a size and a direction. $e^{i θ}$ gave us the direction dial (Step 5); multiplying by $r$ adds the size dial. Together they can hit any point on the sheet.

PICTURE. Same direction, arrow stretched from length $1$ to length $r$ .

The thin arrow is the unit $e^{i θ}$ ; the red arrow is $r$ times longer, landing exactly on $z = x + i y$ .

Step 7 — The degenerate & edge cases (so nothing surprises you)

WHAT. Check the corners where the formula could misbehave — including what a negative angle does.

Worked example Every case, walked

$θ = 0$ : $e^{i 0} = cos 0 + i sin 0 = 1 + 0 i = 1$ . The arrow points straight right. Why fine: zero turn = no rotation.
$θ = π$ (half turn): $e^{iπ} = cos π + i sin π = - 1$ . This is Euler's identity e^(iπ)+1=0. Why fine: a half-turn flips the arrow to point left.
$θ = π /2$ (quarter turn): $e^{iπ /2} = cos \frac{π}{2} + i sin \frac{π}{2} = i$ . Straight up — matches " $i$ = quarter turn" from Step 2.
Negative angle $- θ$ (clockwise): since $cos$ is unchanged by a sign flip but $sin$ flips sign, $e^{- i θ} = cos θ - i sin θ$ . Geometrically this is the arrow turned the other way (clockwise), which is the mirror image across the real axis — exactly the complex conjugate of $e^{i θ}$ . So $e^{- i θ} = \overline{e^{i θ}}$ , and note $e^{i θ} e^{- i θ} = e^{0} = 1$ confirms the length-1 fact from Step 5.
$θ$ and $θ + 2 π$ : give the same point, because a full extra turn lands you where you started. So the argument is only unique up to adding whole turns.
$r = 0$ : $z = 0$ . The origin has no direction — $θ$ is undefined. This is why the parent insists on "nonzero complex number."
"Negative $r$ ": not allowed. $- r e^{i θ} = r e^{i (θ + π)}$ — a negative length is really a positive length pointed the opposite way (half-turn added).

The four coloured landmarks $1, i, - 1, - i$ sit at the four quarter-turns — the same compass as Step 2, now confirmed by the formula. The dashed red arrow shows the clockwise $e^{- i θ}$ as the mirror-image (conjugate) of $e^{i θ}$ .

The one-picture summary

Read it left to right: the sum of Step 1, fed $i θ$ , splits (via the $i$ -cycle, and legally because the series converges absolutely) into the $cos$ pile and the $sin$ pile, which are the legs of a unit arrow; scaling that arrow by $r$ gives $z = r e^{i θ}$ . And because exponents add, turning the dial twice is $e^{i θ_{1}} e^{i θ_{2}} = e^{i (θ_{1} + θ_{2})}$ — the "add the angles" rule that powers Complex multiplication as rotation and scaling, De Moivre's theorem, and Roots of unity.

Recall Feynman: tell the whole story in plain words

We wanted to make sense of " $e$ to an imaginary power." You can't repeat multiplication an imaginary number of times, so we rewrote $e^{u}$ as an endless sum — and because the factorials in the denominators crush the terms, that sum truly homes in on one number for any input. When we poured $i θ$ into that sum, the powers of $i$ marched in a four-beat cycle: right, up, left, down. That cycle sorted the terms into two heaps — a horizontal heap and a vertical heap — and since the sum converges absolutely, we were allowed to reshuffle it into those heaps without changing the answer. The two heaps turned out to be exactly $cos θ$ and $sin θ$ , so $e^{i θ}$ is a horizontal reach of $cos θ$ plus a vertical reach of $sin θ$ : an arrow of length one, tilted by angle $θ$ . Because its two legs obey Pythagoras with hypotenuse one, the arrow never stretches — an imaginary exponent only turns the arrow, while a real exponent would grow it. Turning clockwise ( $- θ$ ) mirrors the arrow across the real axis, giving the conjugate. Finally, multiply that unit arrow by a size $r$ , and you can reach any point on the sheet: $z = r e^{i θ}$ , a stretched, rotated arrow.

Connections

Exponential form z = re^(iθ) (index 3.5.7) — the parent result this page derives visually.
Taylor and Maclaurin series — the sums we relied on in Steps 1 and 4.
Modulus and argument of a complex number — the size $r$ and direction $θ$ .
Polar form (trigonometric) r(cosθ + i sinθ) — the triangle picture of Step 4.
Euler's identity e^(iπ)+1=0 — the $θ = π$ landmark in Step 7.
Complex multiplication as rotation and scaling, De Moivre's theorem, Roots of unity — where "add the angles" pays off.