Complex Roots

LRT04/24/2019

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The complex exponential function

Suppose z = α+ iβ is a complex number with α and β real numbers.Define

ez = eα(cos(β) + isin(β)

)Properties:ez1+z2 = ez1ez2 : the proof follows from the real exponential result andthe trig addition formulas.ez 6= 0 for all z ∈ C: eα is never 0 and the sine and cosine nevervanish for the same β.eα+i0 = eα: the complex exponential agrees with the real exponentialwhen restricted to real numbers.

If z 6= 0 write z =√α2 + β2

(α√

α2 + β2+ i

β√α2 + β2

)=

|z|(cos(arg(z)) + i sin(arg(z))

)with |z| > 0 and 0 6 arg(z) < 2π.

z = eln |z|(cos(arg(z)) + i sin(arg(z))

)= eln |z|+i arg(z)

Hence the exponential function is onto C− {0}. It is not one-to-onesince ez = ez+2kπi for any integer k so there is no continuous complexlog function.

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ez1 = ez2 if and only if z1 − z2 = 2kπi for some integer k.∣∣ez∣∣ = e|z|. ez = ez.

For fixed z, ezt is a function from R to C anddezt

dt= zezt.

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Basis for homogeneous linear 2nd order ODE withconstant coefficients - complex roots

Basis for kerL is y1(t) = e(α+iβ)t, y2(t) = e(α−iβ)t for t ∈ (−∞,∞).It follows that z1y1 + z2y2 ∈ kerL for any constants z1 and z2 ∈ C.

Let

x1(t) =1

2

(y1(t) + y2(t)

)= eαtcos(βt) and let

x2(t) =1

2i

(y1(t)− y2(t)

)= eαtsin(βt).

Since y1(t) = x1(t) + ix2(t) and y2(t) = x1(t)− ix2(t), x1(t) and x2(t)are also a basis and are real-valued functions. They form a basis forthe space of real-valued solutions to the homogeneous equationL(y) = 0.

x′1(t) = eαtαcos(βt)− βeαtsin(βt);

x′1(t) = αx1(t)− βx2(t).

x′2(t) = eαtαsin(βt) + eαtβcos(βt);

x′2(t) = αx2(t) + βx1(t)

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IVP

The initial value problem is easiest to solve when t0 = 0. If you get anon-zero t0 problem solve the same problem with t0 = 0 to get ψ(t)and then your solution is φ(t) = ψ(t− t0).The reason t0 = 0 is so easy is that the Wronskian matrix at 0 is[

1 0α β

]so you want to solve the augmented system

1 0 y0α β y′0

→ 1 0 y00 β y′0 − αy0

ψ(t) = y0x1(t) +y′0 − αy0

βx2(t)

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Example

Solve 3y′′ − y′ + 2y = 0, y(0) = 2, y′(0) = 0.

pL(r) = 3r2 − r + 2; r =1±

√(−1)2 − 4 · 3 · 2

2 · 3=

1

6± i√

23

6.

x1(t) = et6 cos

(√23 t

6

)x2(t) = e

t6 sin

(√23 t

6

)

x′1(t) = et6

1

6cos

(√23 t

6

)− e t

6

√23

6sin

(√23 t

6

)

x′2(t) = et6

1

6sin

(√23 t

6

)+ e

t6

√23

6cos

(√23 t

6

)

Let α =1

6, β =

√23

6.

Then x1(t) = eαtcos(βt) and x2(t) = eαtsin(βt)

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The augmented system is[1 0 2α β 0

]→[1 0 20 β −2α

]→

[1 0 2

0 1 −2α

β

]so the solution is

y(t) = et6

(2cos

(√23 t

6

)− 2√

23sin

(√23 t

6

))

The following is not in the book until section 3.7 to which, alas, weshall not get. However given Acos(ωt) +Bsin(ωt) =√A2 +B2

(A√

A2 +B2cos(ωt) +

B√A2 +B2

sin(ωt)

).

There exists a unique angle δ, 0 6 δ < 2π so that cos(δ) =A√

A2 +B2,

sin(δ) =B√

A2 +B2. Then δ = arccos

(A√

A2 +B2

)if B > 0 and

δ = 2π − arccos

(A√

A2 +B2

)if B < 0.

Acos(ωt) +Bsin(ωt) =√A2 +B2

(cos(ωt− δ)

)7 / 14

Applying this last remark to the solution on the previous slide A = 2,

B = − 2√23

, A2 +B2 = 4 +4

23=

4 · 24

23=

96

23,

√A2 +B2 =

√4 · 24

23= 2

√24

23.

cos(δ) =2

2

√24

23

=

√23

24sin(δ) =

− 2√23

2

√24

23

= − 1√24

.

Hence δ = 2π − arccos

(√23

24

)≈ 6.077616 and

y(t) = 4

√24

23e

t6

(cos

(√23 t

6− δ

))

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The blue curve is y(t) and the red curves are y = ±4

√24

23et/6.

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Same example with different IVP

Solve 3y′′ − y′ + 2y = 0, y(3) = 2, y′(3) = 0.

y(t) = 2

√24

23e

t−36

(2cos

(√23(t− 3)

6

)− 2√

23sin

(√23(t− 3)

6

))or

y(t) = 4

√24

23e

t−36

(cos

(√23 (t− 3)

6− δ

))

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The blue and red curves are before and the green curve is the solutionto the IVP on the last slide.

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Example

Solve 3y′′ + y′ + 2y = 0, y(0) = 2, y′(0) = 0.

pL(r) = 3r2 + r + 2; r =−1±

√(+1)2 − 4 · 3 · 2

2 · 3= −1

6± i√

23

6.

x1(t) = e−t6 cos

(√23 t

6

)x2(t) = e−

t6 sin

(√23 t

6

)

x′1(t) = e−t6

1

6cos

(√23 t

6

)− e− t

6

√23

6sin

(√23 t

6

)

x′2(t) = e−t6

1

6sin

(√23 t

6

)+ e−

t6

√23

6cos

(√23 t

6

)

Let α = −1

6, β =

√23

6.

Then x1(t) = eαtcos(βt) and x2(t) = eαtsin(βt)

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The augmented system is[1 0 2α β 0

]→[1 0 20 β −2α

]→

[1 0 2

0 1 −2α

β

]so the solution is

y(t) = e−t6

(2cos

(√23 t

6

)+

2√23

sin

(√23 t

6

))

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A = 2, B =2√23

, A2 +B2 = 4 +4

23=

4 · 24

23=

96

23,

√A2 +B2 =

√4 · 24

23= 2

√24

23.

cos(δ) =2

2

√24

23

=

√23

24, sin(δ) =

2√23

2

√24

23

=1√24

, and

δ = arccos

(√23

24

)≈ 0.20557.

y(t) = 4

√24

23e−

t6

(cos

(√23 t

6− δ

))

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