This child page takes the parent formula e i θ = cos θ + i sin θ every kind of situation you can meet. Before any example we lay out the full map of cases, so you can see nothing is skipped.
Everything here assumes only two facts, both built in the parent note:
e i θ unit-length clock hand turned by angle θ cos θ sin θ Multiplying two of these hands adds their angles : e i A e i B = e i ( A + B )
Every worked example below is tagged with the cell it fills. The goal: leave no empty cell .
Cell
What can vary / go wrong
Example
C1 Special angle, one quadrantθ π Ex 1
C2 All four quadrantswhich sign do cos θ , sin θ
Ex 2
C3 Degenerate / zero angleθ = 0 θ Ex 3
C4 Negative & large anglesθ < 0 θ > 2 π Ex 4
C5 Convert rectangular → → ( a + bi ) n Ex 5
C6 Trig identity from algebradouble-angle, product rules
Ex 6
C7 Roots (fractional power, many answers)n Ex 7
C8 Real-world word problemrotating physical thing
Ex 8
C9 Exam twist — limiting/growth trape ( a + bi ) θ and growEx 9
Worked example Ex 1 — Evaluatee iπ /3 Forecast: guess before reading — is the answer above or below the real axis? Bigger real part or bigger imaginary part?
Identify θ = π /3 Why this step? π /3 = 60° positive .Apply the formula: e iπ /3 = cos 3 π + i sin 3 π e i θ Read off the standard values: cos 3 π = 2 1 sin 3 π = 2 3 e iπ /3 = 2 1 + i 2 3 . 2 3 ≈ 0.87 > 2 1 steeply upward — matching a 60°
Verify: 2 1 + i 2 3 = 4 1 + 4 3 = 1.
The single biggest source of error is guessing the sign of cos θ sin θ rule : cos θ sin θ
Worked example Ex 2 — Evaluatee i θ each quadrant
Forecast: in which quadrant are BOTH parts negative?
Take θ = 4 π 4 3 π 4 5 π 4 7 π cos , sin = ± 2 1 signs change.
QI, θ = 4 π right & up ⇒ e iπ /4 = 2 1 + 2 i QII, θ = 4 3 π left & up ⇒ cos < 0 , sin > 0 − 2 1 + 2 i QIII, θ = 4 5 π left & down ⇒ − 2 1 − 2 i QIV, θ = 4 7 π right & down ⇒ cos > 0 , sin < 0 2 1 − 2 i
Why these steps? The quadrant fixes the signs — you never memorise the four answers, you read them off the picture (right/left ⇒ real sign, up/down ⇒ imaginary sign).
Verify: each has modulus 2 1 + 2 1 = 1 90°
Worked example Ex 3 — Evaluatee i 0 e iπ e i 2 π Forecast: two of these give the same point. Which two, and why?
θ = 0 e i 0 = cos 0 + i sin 0 = 1 + 0 i = 1 θ = π e iπ = cos π + i sin π = − 1 + 0 i = − 1 180° Euler's identity .θ = 2 π e i 2 π = cos 2 π + i sin 2 π = 1 e i 2 π = e i 0 = 1
Verify: e i 0 = e i 2 π = 1 e iπ = − 1 e iπ + 1 = 0
Common mistake The degenerate trape i 2 π = 1 not 2 π back to 1 ; the number 2 π
Because a full turn is 2 π 2 π same point. Negative θ clockwise .
Worked example Ex 4 — Evaluatee − iπ /2 e i 9 π /2 Forecast: these look different — but they land on the same axis. Guess which.
e − iπ /2 cos ( − 2 π ) + i sin ( − 2 π ) = 0 + i ( − 1 ) = − i bottom .Reduce 2 9 π subtract full turns: 2 9 π − 2 π ⋅ 2 = 2 9 π − 2 8 π = 2 π 2 π θ ∈ [ 0 , 2 π ) So e i 9 π /2 = e iπ /2 = i
Verify: e − iπ /2 ⋅ e iπ /2 = e i 0 = 1 ( − i ) ( i ) = − i 2 = 1 − i i
Worked example Ex 5 — Compute( − 1 + i ) 6 Forecast: the answer is a pure real, pure imaginary, or mixed number — which?
Modulus: r = ( − 1 ) 2 + 1 2 = 2 ∣ z ∣ n ∣ z ∣ Complex Numbers — Polar / Modulus-Argument form ).Argument: − 1 + i QII (left, up), so θ = π − 4 π = 4 3 π arctan ( 1/ − 1 ) = − 4 π Polar form: − 1 + i = 2 e i 3 π /4 Apply De Moivre's Theorem : ( − 1 + i ) 6 = ( 2 ) 6 e i ⋅ 6 ⋅ 3 π /4 = 8 e i 9 π /2 Reduce the angle: 2 9 π ≡ 2 π = 8 e iπ /2 = 8 i 6 [ 0 , 2 π )
Verify: direct multiplication — ( − 1 + i ) 2 = − 2 i ( − 1 + i ) 6 = ( − 2 i ) 3 = − 8 i 3 = − 8 ( − i ) = 8 i
Worked example Ex 6 — Provesin 2 θ = 2 sin θ cos θ Forecast: which part (real or imaginary) of e i 2 θ
Two names for one number: e i 2 θ = cos 2 θ + i sin 2 θ and e i 2 θ = ( e i θ ) 2 = ( cos θ + i sin θ ) 2 Expand the square: ( cos θ + i sin θ ) 2 = cos 2 θ − sin 2 θ + i ( 2 sin θ cos θ ) i 2 = − 1 sin 2 Match imaginary parts: sin 2 θ = 2 sin θ cos θ cos 2 θ = cos 2 θ − sin 2 θ
Verify: at θ = 6 π sin 3 π = 2 3 2 ⋅ 2 1 ⋅ 2 3 = 2 3
A whole power gives one answer; a root gives several , spaced evenly around the circle. This is the Roots of Unity idea. The figure shows why: adding 2 π k
Worked example Ex 7 — Find all cube roots of8 i Forecast: how many answers, and how far apart in angle?
Polar form of 8 i modulus 8 2 π 8 i = 8 e i ( π /2 + 2 π k ) + 2 π k 8 i Take cube roots: ( 8 i ) 1/3 = 8 1/3 e i ( π /2 + 2 π k ) /3 = 2 e i ( π /6 + 2 π k /3 ) 8 1/3 = 2 Sweep k = 0 , 1 , 2 (then it repeats):
k = 0 : 2 e iπ /6 = 2 ( 2 3 + 2 1 i ) = 3 + i . k = 1 : 2 e i 5 π /6 = 2 ( − 2 3 + 2 1 i ) = − 3 + i . k = 2 : 2 e i 3 π /2 = 2 ( 0 − i ) = − 2 i . 3 2 π = 120°
Verify: cube the first, ( 3 + i ) 3 = 8 i 2 e iπ /6 = 8 e iπ /2 = 8 i 0 ( 3 + i ) + ( − 3 + i ) + ( − 2 i ) = 0
Worked example Ex 8 — A rotating spokeA wheel of radius 1 m ( 1 , 0 ) 2 π Where is the dot after 3
Forecast: three quarter-turns — will the dot be on an axis or between axes?
Angle turned: θ = 2 π × 3 = 2 3 π × s rad × s = rad Model position as e i θ the dot is at e i 3 π /2 = cos 2 3 π + i sin 2 3 π = 0 − i = − i is e i θ x y Translate to coordinates: − i x = 0 , y = − 1 ( 0 , − 1 )
Verify: three quarter-turns from the right side ( 1 , 0 ) bottom . Consistent. Also ∣ − i ∣ = 1
The parent's Trap A warns that e i θ real part too ? Then you get spinning and growing — a spiral.
Worked example Ex 9 — Evaluatee ( 1 + i ) π Forecast: will the modulus be 1 1
Split the exponent: e ( 1 + i ) π = e π ⋅ e iπ sum in the exponent into a product : e a + b = e a e b Rotation part: e iπ = − 1 Scaling part: e π ≈ 23.14 real exponent grows — this is the growth Trap A warned you not to attach to a pure imaginary exponent.Combine: e ( 1 + i ) π = e π ⋅ ( − 1 ) = − e π ≈ − 23.14
Verify: ∣ e ( 1 + i ) π ∣ = e π ≈ 23.14 = 1 off the unit circle, exactly because the exponent had a nonzero real part. The pure-imaginary case e iπ = − 1 1
Recall Which cell does each example fill?Ex 1 → C1 (special angle QI) ::: Ex 2 → C2 (all four quadrants)
Ex 3 → C3 (degenerate: 0 , π , 2 π
