4.02n - Euler's formula

4.02n - Euler's formula

Use Euler's formula to connect the complex exponential with cosine and sine. You will interpret an imaginary exponent as an angle on the unit circle, calculate exact values, and solve equations by identifying a direction and its full turns.

Imaginary exponentials on the unit circle

What is eiπe^{i\pi}? The real exponential eπe^\pi is larger than 2020, so an exponent involving π\pi might suggest a large answer. Euler's formula instead puts eiπe^{i\pi} at the ordinary number −1-1. The imaginary exponent records a turn, rather than growth along the positive real axis.

You already know

The point cos⁡θ+isin⁡θ\cos\theta + i\sin\theta has modulus 11 and argument θ\theta. Multiplying two such numbers adds their angles, by the compound-angle identities. Radians measure a full turn as 2π2\pi.

The expression eiθe^{i\theta}, for real θ\theta, is the complex exponential on the unit circle. Euler's formula tells you exactly what its two coordinates are. You do not multiply eθe^\theta by ii: the ii is part of the exponent.

[DIAGRAM: asset_name: Euler's formula on the unit circle; asset_slug: h245_4_02_euler_circle; caption: The real and imaginary parts of an imaginary exponential are cosine and sine; description: Equal-scale Argand axes with a circle of radius 1 and point e to the i theta at theta = pi/3. A blue position vector, dashed coordinate guides, and a labelled angle theta show coordinates cos theta and sin theta. The four axis points are labelled e to the 0 = 1, e to the i pi/2 = i, e to the i pi = -1 and e to the -i pi/2 = -i.]
The real and imaginary parts of an imaginary exponential are cosine and sine

Euler's formula

For real θ\theta, measured in radians:

eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta

Consequently, Re(eiθ)=cos⁡θ\mathrm{Re}(e^{i\theta}) = \cos\theta, Im(eiθ)=sin⁡θ\mathrm{Im}(e^{i\theta}) = \sin\theta, and ∣eiθ∣=1|e^{i\theta}| = 1.
Not in the formula booklet: learn it.

The modulus statement follows from cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1. As θ\theta increases, the point travels anticlockwise round the circle. A negative θ\theta travels clockwise. At θ=0\theta = 0, e0=1e^0 = 1; after a half turn, eiπ=−1e^{i\pi} = -1.

The exponential notation agrees with the angle-addition rule:

eiαeiβ=(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)=cos⁡(α+β)+isin⁡(α+β)=ei(α+β)e^{i\alpha}e^{i\beta} = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta) = e^{i(\alpha + \beta)}

The circle's addition of angles is the exponential's addition of exponents.

Here is the exact surprising identity behind the opening question:

eiπ+1=cos⁡π+isin⁡π+1=−1+0i+1=0e^{i\pi} + 1 = \cos\pi + i\sin\pi + 1 = -1 + 0i + 1 = 0

The constants ee, ii, π\pi, 11 and 00 meet in one equation because the exponential describes the geometry of turning.

Key idea

For real θ\theta, eiθe^{i\theta} has unit modulus and direction θ\theta. Its real and imaginary coordinates are cosine and sine.

Explain the two coordinates as well as the length: saying only “it is a rotation” leaves the formula's content unstated.

Exact values from Euler's formula

To evaluate an imaginary exponential, replace it by its cosine and sine, keeping the angle unchanged in both functions. Exact trigonometric values then give cartesian form.

Worked example 1: a positive angle

Question. Express eiπ/3e^{i\pi/3} in the form a+bia + bi.

Plan. Apply Euler's formula and use exact values at π/3\pi/3.

eiπ/3=cos⁡π3+isin⁡π3Euler’s formula=12+32iexact trigonometric values\begin{aligned} e^{i\pi/3} &= \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} && \text{Euler's formula} \\ &= \frac12 + \frac{\sqrt3}{2}i && \text{exact trigonometric values} \end{aligned}

Why this step? The exponent's real angle is π/3\pi/3; that angle supplies both coordinates.

Answer. 12+32i\frac12 + \frac{\sqrt3}{2}i.

Check. Its modulus is 1/4+3/4=1\sqrt{1/4 + 3/4} = 1, and both parts are positive, as expected in the first quadrant.

Change only the sign of the angle. The real part stays the same and the imaginary part changes sign, so the value is the conjugate.

Worked example 2: a negative angle

Question. Express e−iπ/3e^{-i\pi/3} in the form a+bia + bi.

Plan. Use Euler's formula with angle −π/3-\pi/3, then use the evenness of cosine and oddness of sine.

e−iπ/3=cos⁡(−π3)+isin⁡(−π3)Euler’s formula=cos⁡π3−isin⁡π3even cosine, odd sine=12−32iexact values\begin{aligned} e^{-i\pi/3} &= \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right) && \text{Euler's formula} \\ &= \cos\frac{\pi}{3} - i\sin\frac{\pi}{3} && \text{even cosine, odd sine} \\ &= \frac12 - \frac{\sqrt3}{2}i && \text{exact values} \end{aligned}

Why this step? Clockwise and anticlockwise angles reflect the point in the real axis, which reverses only the imaginary coordinate.

Answer. 12−32i\frac12 - \frac{\sqrt3}{2}i.

Check. Multiplying the two example answers gives 1/4+3/4=11/4 + 3/4 = 1, matching eiπ/3e−iπ/3=e0e^{i\pi/3}e^{-i\pi/3} = e^0.

The conjugation rule (eiθ)∗=e−iθ(e^{i\theta})^* = e^{-i\theta} follows directly from Euler's formula and is not in the formula booklet: learn it.

Watch out

Wrong: e−iθ=−cos⁡θ−isin⁡θe^{-i\theta} = -\cos\theta - i\sin\theta. A negative exponent changes the angle, not the sign of the whole number. Right: e−iθ=cos⁡θ−isin⁡θe^{-i\theta} = \cos\theta - i\sin\theta.

Use the same method for a different clockwise turn.

Periodicity and equal exponentials

Different imaginary exponents can represent the same point, because a full turn changes neither cosine nor sine. Unlike the real exponential, the imaginary exponential repeats.

For example, e0i=1e^{0i} = 1 and e2πi=1e^{2\pi i} = 1, although 0≠2π0 \ne 2\pi. You cannot cancel exponentials and claim their imaginary exponents must be equal.

Equal imaginary exponentials

For real α\alpha, β\beta:

eiα=eiβ⟺α=β+2kπ for some k∈Ze^{i\alpha} = e^{i\beta} \quad \Longleftrightarrow \quad \alpha = \beta + 2k\pi \text{ for some } k \in \mathbb{Z}

The angles describe the same direction exactly when they differ by a whole number of full turns.
Not in the formula booklet: learn it.

When solving for an angle, first identify one direction, then include the full turns, then apply any interval restriction.

Worked example 3: all angles for one point

Question. Determine all real θ\theta such that eiθ=ie^{i\theta} = i.

Plan. Locate ii on the unit circle and add integer multiples of 2π2\pi.

i=cos⁡π2+isin⁡π2positive imaginary axis=eiπ/2Euler’s formulaθ=π2+2kπ,k∈Zall full turns\begin{aligned} i &= \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} && \text{positive imaginary axis} \\ &= e^{i\pi/2} && \text{Euler's formula} \\ \theta &= \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} && \text{all full turns} \end{aligned}

Why this step? A single principal argument supplies one solution; “all real” requires every full-turn equivalent too.

Answer. θ=π/2+2kπ\theta = \pi/2 + 2k\pi, k∈Zk \in \mathbb{Z}.

Check. Adding 2kπ2k\pi leaves cosine 00 and sine 11.

Change only the point from ii to −i-i. The direction reverses, while the full-turn spacing remains 2π2\pi.

Worked example 4: the opposite imaginary point

Question. Determine all real θ\theta such that eiθ=−ie^{i\theta} = -i, and identify the solution in [0,2π)[0, 2\pi).

Plan. Use a clockwise quarter turn, add full turns, then select the required interval.

−i=e−iπ/2negative imaginary axisθ=−π2+2kπ,k∈Zall equivalent anglesθ=3π2the value in [0,2π)\begin{aligned} -i &= e^{-i\pi/2} && \text{negative imaginary axis} \\ \theta &= -\frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} && \text{all equivalent angles} \\ \theta &= \frac{3\pi}{2} && \text{the value in } [0, 2\pi) \end{aligned}

Why this step? The principal interval selects a representative; it does not change the set of directions represented by the exponential.

Answer. All solutions are −π/2+2kπ-\pi/2 + 2k\pi; the one in [0,2π)[0, 2\pi) is 3π/23\pi/2.

Check. cos⁡(3π/2)=0\cos(3\pi/2) = 0, sin⁡(3π/2)=−1\sin(3\pi/2) = -1, giving −i-i.

This periodicity explains why the opening answer is compatible with a whole family: ei(π+2kπ)=−1e^{i(\pi + 2k\pi)} = -1 for every integer kk.

Watch out

Wrong: from eiα=eiβe^{i\alpha} = e^{i\beta}, concluding α=β\alpha = \beta. Right: the angles can differ by 2kπ2k\pi. For example, eiπ=e−iπ=−1e^{i\pi} = e^{-i\pi} = -1.

Make the interval restriction your last step.

Quick Recap

  • Euler's formula is eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta, with θ\theta in radians.
  • The point has modulus 11, real part cos⁡θ\cos\theta, and imaginary part sin⁡θ\sin\theta.
  • Reversing the angle takes the conjugate: e−iθ=(eiθ)∗e^{-i\theta} = (e^{i\theta})^*.
  • Equal imaginary exponentials have angles differing by 2kπ2k\pi; apply the requested interval at the end.

The opening value is exactly eiπ=−1e^{i\pi} = -1, the point reached after a half turn round the unit circle.