Case 3 - complex conjugate roots — Euler's formula connection: Question bank

Sharpen your instincts. Every item below targets a place where students feel right but are wrong, or a boundary case the formula quietly depends on. Cover the parent Case 3 - complex conjugate roots — Euler's formula connection first, then test yourself here.

Figure — Case 3 - complex conjugate roots — Euler's formula connection

True or false — justify

A real-coefficient quadratic can have exactly one complex root and one real root.

False. Conjugation fixes real coefficients, so if

α + i β

is a root then

α - i β

must be too — complex roots always arrive as a pair, never solo.

Δ = b^{2} - 4 a c = 0

you are still in Case 3.

False.

Δ = 0

is Case 2 (a repeated real root, Case 2 — repeated real roots); Case 3 needs

Δ < 0

strictly, so that

b^{2} - 4 a c = i 4 a c - b^{2}

is genuinely imaginary and

β \neq = 0

The general solution

y = e^{α x} (C_{1} cos β x + C_{2} sin β x)

contains only two arbitrary constants even though the ODE has "four" functions

e^{α x}, cos, sin

True. The envelope

e^{α x}

is a common factor, not a free choice; a second-order ODE has exactly two free constants, matching

C_{1}, C_{2}

e^{i β x}

makes the amplitude grow as

x

increases.

False.

∣ e^{i β x} ∣ = cos^{2} β x + sin^{2} β x = 1 = 1

for all

x

— it's a pure rotation on the unit circle, so it never scales amplitude. Only

α

(the real part) touches amplitude.

Choosing

β < 0

instead of

β > 0

gives a genuinely different solution set.

False.

cos

is even and

sin

is odd, so flipping

β

's sign only swaps

C_{2} \to - C_{2}

; since

C_{2}

ranges over all reals anyway, the solution space is identical.

For

α = 0

the solution is periodic with period

2 π / β

True. With no envelope,

y = C_{1} cos β x + C_{2} sin β x

repeats every time

β x

advances by

2 π

, i.e. every

Δ x = 2 π / β

The two complex solutions

e^{(α + i β) x}

and

e^{(α - i β) x}

are linearly independent over the complex numbers.

True. They are distinct exponentials with distinct exponents; their Wronskian is nonzero, so they are independent — we recombine them for realness, not because they were dependent.

Spot the error

(Recall: $m$ is a characteristic root, i.e. a solution of $a m^{2} + bm + c = 0$ , written $m = α \pm i β$ .)

m = - 1 \pm 2 i

, so the envelope is

e^{(- 1 + 2 i) x}

and it decays and oscillates from the single exponential."

The envelope is only

e^{α x} = e^{- x}

; the

2 i

lives in

e^{2 i x} = cos 2 x + i sin 2 x

, which oscillates but has magnitude

1

. Splitting

e^{α x}

(size) from

e^{i β x}

(rotation) is the whole point of Euler here.

"Roots are

\pm 3 i

, so

y = e^{3 i x} (C_{1} cos + C_{2} sin)

Double error:

α = 0

means the envelope is

e^{0} = 1

(no

e^{3 i x}

out front), and you never leave an

i

in a real general solution. The answer is

y = C_{1} cos 3 x + C_{2} sin 3 x

"For

y^{''} + 2 y^{'} + 5 y = 0

I get

m = - 1 \pm 2 i

, so

y = C_{1} cos 2 x + C_{2} sin 2 x

The

e^{α x} = e^{- x}

envelope was dropped. Since

α = - 1 \neq = 0

, the correct answer is

y = e^{- x} (C_{1} cos 2 x + C_{2} sin 2 x)

; forgetting the envelope loses the decay entirely.

"Euler's formula is a lucky coincidence — no reason

e^{i θ}

should equal

cos + i sin

Not luck: define

e^{i θ}

by its Maclaurin series and the powers of

i

cycle

1, i, - 1, - i

, splitting the sum exactly into the cosine (even) and

i sin

(odd) series. It's forced, not coincidental.

"To get real solutions I add

y_{1} + y_{2}

; that's my only real solution."

You get two real solutions:

u = \frac{1}{2} (y_{1} + y_{2}) = e^{α x} cos β x

and

v = \frac{1}{2 i} (y_{1} - y_{2}) = e^{α x} sin β x

. Both are needed for a two-dimensional solution space.

"Since the complex form

C_{1} e^{(α + i β) x} + C_{2} e^{(α - i β) x}

has an

i

in it, it's wrong."

It's actually a valid general solution — but only if you let

C_{1}, C_{2}

be complex conjugates so the imaginary parts cancel. The real form is preferred because it makes realness manifest without constraints on the constants.

Why questions

Why do we appeal to the superposition principle before recombining

y_{1}, y_{2}

Because the ODE is linear and homogeneous, any linear combination of solutions is again a solution — that licence lets us form

u

and

v

, which are combinations chosen precisely to kill the imaginary parts.

Why divide by

2 i

(not

2

) when building

v = \frac{1}{2 i} (y_{1} - y_{2})

The difference

y_{1} - y_{2}

equals

2 i e^{α x} sin β x

, so dividing by

2 i

strips the

i

and the factor

2

, leaving the clean real function

e^{α x} sin β x

Why must we check the Wronskian rather than just eyeballing

u

and

v

To guarantee the two solutions span the whole solution space;

W = β e^{2 α x}

is nonzero (since

β \neq = 0

e^{2 α x} > 0

), proving independence rigorously rather than by appearance.

Why does

α = - \frac{b}{2 a}

decide growth versus decay?

α

is the exponent of the real envelope

e^{α x}

α < 0

shrinks amplitude (damped),

α > 0

grows it,

α = 0

holds it constant — this is the friction term in a damped oscillator.

Why can't Case 3's method (Euler) be replaced by Case 1's method of two real exponentials?

Because with

Δ < 0

the "exponents"

α \pm i β

are not real, so

e^{(α \pm i β) x}

are complex-valued — Case 1 — real distinct roots never needs Euler because its exponents stay real.

Why is

e^{i β x}

best pictured as a rotating arrow on the unit circle?

Because

cos β x

and

sin β x

are its horizontal and vertical shadows, and their squares sum to

1

; as

x

grows the arrow spins at rate

β

, giving the oscillation without ever changing length.

Why does a negative discriminant force oscillation physically?

A negative

Δ

means the restoring force dominates the damping, so the system overshoots and swings back repeatedly — the imaginary part

β

is exactly that swing's angular frequency.

Complex roots $\Rightarrow$ $α = - \frac{b}{2 a}$ is the envelope (real part, decides growth/decay), $β = \frac{4 a c - b ^{2}}{2 a}$ is the wiggle (imaginary part, decides frequency), and Euler is the bridge that turns the scary $i$ into ordinary sines and cosines.

Question bank — Case 3 - complex conjugate roots — Euler's formula connection

True or false — justify

Spot the error

Why questions

Edge cases