Differential-Equations · Unit 6 · Video 2 · Interactive Practice

Euler's e^(iθ) Passes Both Exponential Tests: Multiply the Moduli, Add the Angles

IKey Formulas

Formula	Name	What it gives
$e^{a+ib} = e^{a}\,e^{ib} = e^{a}(\cos b + i\sin b)$	Complex exponential (a definition)	With $a = 0$ it is Euler's $e^{i\theta} = \cos\theta + i\sin\theta$ ; $e^a$ is a positive real number
$e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1+\theta_2)}$	Exponential law (test 1)	The addition formulas for $\cos(\theta_1+\theta_2)$ and $\sin(\theta_1+\theta_2)$ in one line
$\dfrac{d}{dt}e^{it} = i\,e^{it}, \quad e^{i\cdot 0} = 1$	Differential equation and initial value (test 2)	Checked with $(u + iv)' = u' + iv'$ and with $\cos 0 + i\sin 0 = 1$
$r_1e^{i\theta_1}\cdot r_2e^{i\theta_2} = r_1r_2\,e^{i(\theta_1+\theta_2)}$	Polar product	Multiply the moduli, add the arguments

Key Insight: A definition earns the name exponential by behaving like one. $\cos\theta + i\sin\theta$ obeys the exponential law and solves $y' = iy$ , $y(0) = 1$ , so it may be written $e^{i\theta}$ , and then the polar form $\alpha = re^{i\theta}$ turns multiplication into a stretch by $r$ and a rotation by $\theta$ .

IIMultiply the Moduli, Add the Angles

The product $\alpha_1\alpha_2$ has length $r_1r_2$ and angle $\theta_1 + \theta_2$ , whatever the two factors are.

💡 Challenge: with $|\alpha_1|$ at least about $0.6$ , place $\alpha_2$ so that the product lands on $1$ . The point you have built is $\alpha_1^{-1} = \tfrac{1}{r_1}e^{-i\theta_1}$ : reciprocal modulus, opposite angle.

IIIThe Derivative of $e^{it}$ Is $i\,e^{it}$

Differentiating $\cos t + i\sin t$ part by part gives a velocity equal to the position turned a quarter turn.

t 0.80

IV $e^{a+ib}$ : $a$ Sets the Length, $b$ Sets the Angle

In $e^{a+ib} = e^a(\cos b + i\sin b)$ the real part of the exponent scales, the imaginary part rotates.

💡 Because $b$ enters only through $\cos b$ and $\sin b$ , the exponents $z$ and $z + 2\pi i$ give the same value: $e^{2\pi i} = 1$ . And since $e^a > 0$ , the value $e^{a+ib}$ is never $0$ .

VQuiz Questions

Problem 1 · A Polar Product

Given: $\alpha = 2e^{i\pi/6}$ and $\beta = 3e^{i\pi/3}$ — find the product $\alpha\beta$ .

✅ Correct! Modulus

2 \cdot 3 = 6

, argument

\tfrac{\pi}{6} + \tfrac{\pi}{3} = \tfrac{\pi}{2}

, and

6e^{i\pi/2} = 6i

❌ Not quite. The angle

\tfrac{\pi}{2}

is right, but moduli multiply:

r_1 r_2 = 6

, not

r_1 + r_2

❌ Not quite. The moduli multiply, but the angles add: the exponential law gives

e^{i\pi/6}e^{i\pi/3} = e^{i(\pi/6 + \pi/3)}

❌ Not quite. This is

6e^{i\pi/3}

: the modulus is right, but the angle of

\alpha

was dropped. The arguments add to

\tfrac{\pi}{2}

❌ Not quite. Multiply the moduli and add the angles.

Show solution

Multiply as ordinary products and combine the exponentials with the exponential law:

\alpha\beta = (2 \cdot 3)\,e^{i\pi/6}e^{i\pi/3} = 6\,e^{i(\pi/6 + \pi/3)} = 6\,e^{i\pi/2}.

By Euler's definition, $e^{i\pi/2} = \cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} = 0 + i = i$ , so $\alpha\beta = 6i$ .

Check in Cartesian form: $\alpha = \sqrt{3} + i$ and $\beta = \tfrac{3}{2} + \tfrac{3\sqrt{3}}{2}i$ , so $(ac - bd) + i(ad + bc) = \big(\tfrac{3\sqrt3}{2} - \tfrac{3\sqrt3}{2}\big) + i\big(\tfrac{9}{2} + \tfrac{3}{2}\big) = 6i$ .

Problem 2 · Factoring Out the i

Given: $\dfrac{d}{dt}(\cos t + i\sin t) = -\sin t + i\cos t$ — which expression is this equal to?

✅ Correct!

i(\cos t + i\sin t) = i\cos t + i^2\sin t = -\sin t + i\cos t

, so

e^{it}

passes

y' = iy

❌ Not quite.

e^{it} = \cos t + i\sin t

, and its real part

\cos t

does not match

-\sin t

. The derivative carries the factor

i

from the exponent.

❌ Check the sign of $i^2$ .

-i(\cos t + i\sin t) = -i\cos t - i^2\sin t = \sin t - i\cos t

, the negative of the derivative.

❌ Not quite.

e^{-it} = \cos t - i\sin t

, so

i\,e^{-it} = \sin t + i\cos t

: the real part has the wrong sign.

❌ Not quite. Expand each option with Euler's definition and compare real and imaginary parts.

Show solution

Differentiate the real and imaginary parts separately: $(\cos t)' = -\sin t$ and $(\sin t)' = \cos t$ , so the derivative is $-\sin t + i\cos t$ .

Factor out $i$ . The term $-\sin t$ must come from $i \cdot i\sin t = i^2\sin t = -\sin t$ , and $i\cos t$ comes from $i \cdot \cos t$ :

-\sin t + i\cos t = i(\cos t + i\sin t) = i\,e^{it}.

Together with $e^{i\cdot 0} = \cos 0 + i\sin 0 = 1$ , the function $e^{it}$ solves $y' = iy$ , $y(0) = 1$ , exactly as a real exponential $e^{kt}$ solves $y' = ky$ , $y(0) = 1$ .

Problem 3 · A Complex Exponent

Given: $z = e^{\ln 2 + 2\pi i/3}$ — find its modulus $|z|$ and write $z$ in Cartesian form.

What is $|z|$ ?

What is $z$ in the form $x + iy$ ?

✅ Correct!

z = e^{\ln 2}e^{2\pi i/3} = 2\big(-\tfrac12 + \tfrac{\sqrt3}{2}i\big) = -1 + \sqrt3\,i

, at distance

2

and angle

120\degree

❌ Check the modulus. By definition

e^{a+ib} = e^a e^{ib}

with

a = \ln 2

; the factor

e^{ib}

has modulus

1

, so

|z| = e^{\ln 2}

❌ Check the cosine.

\tfrac{2\pi}{3}

is in the second quadrant, so

\cos\tfrac{2\pi}{3} = -\tfrac12

, not

+\tfrac12

❌ Almost. That is

e^{2\pi i/3}

alone; the factor

e^{\ln 2} = 2

still multiplies it.

❌ Check which part is which. The real part is

2\cos\tfrac{2\pi}{3}

and the imaginary part is

2\sin\tfrac{2\pi}{3}

, not the other way round.

❌ Not quite. Compute

2\big(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\big)

Show solution

With $a = \ln 2$ and $b = \tfrac{2\pi}{3}$ , the definition gives

z = e^{\ln 2}\,e^{2\pi i/3} = 2\Big(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\Big).

The first factor is the positive real number $e^{\ln 2} = 2$ , and $|e^{ib}| = \sqrt{\cos^2 b + \sin^2 b} = 1$ , so $|z| = 2$ and $\arg z = \tfrac{2\pi}{3}$ .

z = 2\Big(-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}\,i\Big) = -1 + \sqrt{3}\,i.

Check: $|-1 + \sqrt3\,i| = \sqrt{1 + 3} = 2$ .

Problem 4 · A Triple-Angle Formula from the Exponential Law

Given: the exponential law gives $e^{3i\theta} = \big(e^{i\theta}\big)^3$ , that is $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$ — find $\cos 3\theta$ by comparing real parts.

✅ Correct! The real terms of the expansion are

\cos^3\theta

and

3\cos\theta\,(i\sin\theta)^2 = -3\cos\theta\sin^2\theta

❌ Check the factor $i^2$ . The term

3\cos\theta\,(i\sin\theta)^2

carries

i^2 = -1

, so it enters with a minus sign.

❌ Not quite. A cube of a sum is not the sum of the cubes;

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

. Also

(i\sin\theta)^3 = -i\sin^3\theta

is imaginary.

❌ That is the other half. These terms carry a factor

i

: they give

\sin 3\theta

, not

\cos 3\theta

❌ Not quite. Expand

(\cos\theta + i\sin\theta)^3

binomially and keep the terms with no

i

Show solution

Write $c = \cos\theta$ , $s = \sin\theta$ and expand with the binomial theorem:

(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2s - 3cs^2 - is^3.

Group the terms using $i^2 = -1$ and $i^3 = -i$ :

(c + is)^3 = \big(c^3 - 3cs^2\big) + i\big(3c^2s - s^3\big).

The left side is $\cos 3\theta + i\sin 3\theta$ , and two complex numbers are equal exactly when their real parts and their imaginary parts agree:

\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta, \qquad \sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta.

Check at $\theta = \tfrac{\pi}{3}$ : $\cos\pi = -1$ and $\tfrac18 - 3\cdot\tfrac12\cdot\tfrac34 = \tfrac18 - \tfrac98 = -1$ . As in the video, one line of the exponential law packs two trigonometric identities.

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Euler's e^(iθ) Passes Both Exponential Tests: Multiply the Moduli, Add the Angles

IKey Formulas

IIMultiply the Moduli, Add the Angles

IIIThe Derivative of eite^{it}eit Is i eiti\,e^{it}ieit

IVea+ibe^{a+ib}ea+ib: aaa Sets the Length, bbb Sets the Angle

VQuiz Questions

IIIThe Derivative of $e^{it}$ Is $i\,e^{it}$

IV $e^{a+ib}$ : $a$ Sets the Length, $b$ Sets the Angle