Euler’s Formula (Class Notes from Feb. 11, 2013)

Yoichiro Mori

February 13, 2013

Recall the Taylor’s formula for the exponential, sine and cosine functions:

exp(x) = 1 + x+x2

2!+ · · ·+ xn

n!+ · · ·

sin(x) = x− x3

3!+

x5

5!· · ·

cos(x) = 1− x2

2!+

x4

4!· · ·

The above formulas are valid for all real numbers, as you have most probablylearned in calculus II.

The above equations are the Taylor series of familiar functions exp(x),sin(x) and cos(x). But once established, it is also possible to see the aboveexpressions as definitions of these functions. The advantage of this is thatthe infinite series makes sense not only for real numbers but also for complexnumbers (and, as we shall see later on, also for matrices).

Substitute x = it(where i is the imaginary unit and t is a real number)in the above Taylor expansion. We find:

exp(it) = 1 + it+(it)2

2+

(it)3

3!+

(it)4

4!+

(it)5

5!· · ·

= 1− t2

2!+

t4

4!· · ·+ i

(

t− t3

3!+

t5

5!+ · · ·

)

= cos(t) + i sin(t).

(1)

We thus have Euler’s formula:

exp(it) = cos(t) + i sin(t). (2)

In general, if we have exp(a+ ib) where a and b are real numbers, we have:

exp(a)(cos(b) + i sin(b)). (3)

1

Some useful expressions:

sin(t) =exp(it)− exp(−it)

2i, cos(t) =

exp(it) + exp(−it)

2. (4)

For a complex number α = a+ ib, a, b ∈ R, define

Reα = a, Imα = b. (5)

Using these expressions, we have:

Re exp(it) = cos(t), Im exp(it) = sin(t). (6)

Let us investigate the action of differentiation on exp(it). If we differen-tiate this equation with respect to t, once, and then twice, we have:

d

dtexp(it) = i exp(it),

d2

dt2exp(it) = i2 exp(it) = − exp(it). (7)

Looking at the second derivative and using Euler’s formula, we have:

d2

dt2(cos(t) + i sin(t)) = − (cos(t) + i sin(t)) . (8)

This is nothing other than the statement that both the sine and cosinefunction acquire a negative sign upon differentiating twice. Through Euler’sformula, this fact is linked to the algebraic relation i2 = −1. Don’t youthink this is a beautiful observation! This observation is not just beautiful.In fact, it will be foundational in our discussion of finding solution to secondorder, higher order, and systems of linear differential equations.

One reason why Euler’s formula is very useful is that the algebra andcalculus of exponential functions is simpler than trigonometric functions.For trigonometric functions, the addition formula (that is cos(a+b) = · · · ) isquite complicated involving sines and cosines, whereas the addition formulafor exponential functions , exp(a+ b) = exp(a) exp(b), is exceedingly simple.For exponential functions, differentiation and integration is the same asmultiplication by a constant, whereas, for trigonometric functions, there isthe additional complication of sine changing to cosine and vice versa. Let uslook at examples of this. You may recall the following triple angle formulafrom trigonometry:

cos(3θ) = 4 cos3(θ)− 3 cos(θ). (9)

MATH 2574H 2 Yoichiro Mori

Let us verify this formula using Euler’s formula. Let us compute cos3(θ).

cos3(θ) =

(

exp(iθ) + exp(−iθ)

2

)3

=1

8(exp(3iθ) + 3 exp(iθ) + 3 exp(−iθ) + exp(−3iθ))

=1

4

(

exp(3iθ) + exp(−3iθ)

2

)

+3

4

(

exp(iθ) + exp(−iθ)

2

)

=1

4cos(3θ) +

3

4cos(θ).

(10)

This is nothing other than the triple angle formula. To derive this, allwe have used is some very basic algebra. In fact, some of you will haverecognized that this method of deriving the triple angle formula generalizesto deriving the n-angle formula for any integer n.

Here is another example. Consider the integral:∫

exp(ax) cos(bx)dx, a 6= 0, b 6= 0. (11)

To refresh our memory, let us find the above indefinite integral in the “usual”way, using integration by parts:

∫

exp(ax) cos(bx)dx =1

aexp(ax) cos(bx) +

∫

b

aexp(ax) sin(bx)dx

=1

aexp(ax) cos(bx) +

b

a2exp(ax) sin(bx)−

∫

b2

a2exp(ax) cos(bx)dx.

(12)

From this, we see that:∫

exp(ax) cos(bx)dx =a

a2 + b2exp(ax) cos(bx)+

b

a2 + b2exp(ax) sin(bx)+C,

(13)where C is the constant of integration. Let us now use Euler’s formula tofind the indefinite integral.

∫

exp(ax) cos(bx)dx =

∫

Re (exp((a+ ib)x)) dx

=Re

∫

exp((a+ ib)x)dx = Re

(

1

a+ ibexp((a+ ib)x)

)

+ C

(14)

The rest is just a bit of algebra to see that the last expression indeed gives(13). No integration by parts, and we get the answer directly.

MATH 2574H 3 Yoichiro Mori

Homework Problems (Due Feb. 19, 2013)

1. Compute the following.

exp(2πi), exp(−πi/2), exp(πi),√2 exp(−πi/4), 2 exp(πi/6).

2. Write the following complex numbers in the form r exp(iθ) where rand θ are real numbers.

i, 1 +√3i, 1 + i, −1 + i, 3 + 4i.

3. Derive the triple angle formula for the sine function imitating thederivation of the triple angle formula for the cosine in (9).

4. Consider the integral:

∫

2π

0

cos(mx) cos(nx)dx, where m,n ∈ Z (15)

(a) Write cos(mx) and cos(nx) in terms of complex exponentials, andcompute cos(mx) cos(nx).

(b) Show that, for integer ℓ,

∫

2π

0

exp(iℓx)dx =

{

2π if ℓ = 0,

0 otherwise.

(c) Using the above, compute (15).

5. Consider the sum:n∑

k=1

cos(kθ). (16)

(a) Show that the above sum is equal to:

Ren∑

k=1

exp(ikθ). (17)

(b) Using the above, compute (16). (Hint: Note exp(ikθ) = (exp(iθ))k.Use summation formula for geometric series).

MATH 2574H 4 Yoichiro Mori