Differential-Equations · Unit 10 · Video 1 · Interactive Practice

One Complex Root Is Enough: The Conjugate Pair and the Complex Form of Real Solutions

IKey Formulas

Formula	Name	Role
$e^{(a-bi)t} = e^{at}\cos bt - i\,e^{at}\sin bt$	Conjugate root	Real part $y_1$ , imaginary part $-y_2$ : nothing new
$y = C_1 e^{(a+bi)t} + C_2 e^{(a-bi)t}$ is real $\iff C_2 = \overline{C_1}$	Complex form	Replace $i$ by $-i$ , then match the coefficients of the two exponentials
$\cos\alpha = \dfrac{e^{i\alpha} + e^{-i\alpha}}{2},\quad \sin\alpha = \dfrac{e^{i\alpha} - e^{-i\alpha}}{2i}$	Euler read backwards	The sine formula has $2i$ , not $2$ , in the denominator
$C_1 = c + id \;\Rightarrow\; y = e^{at}(2c\cos bt - 2d\sin bt)$	Back to real form	$c_1 = 2c$ , $c_2 = -2d$ ; conversely $c = c_1/2$ , $d = -c_2/2$

Key Insight: One complex root is enough. The root $a + bi$ already gives $y_1$ and $y_2$ ; in the complex form the second term must be the conjugate of the first, so the real solutions carry exactly two real constants, $c$ and $d$ .

IIThe Conjugate Root Adds Nothing New

For $y'' + 2y' + 5y = 0$ , with roots $-1 \pm 2i$ , the conjugate root's exponential keeps the real part and negates the imaginary part.

Root t 0.60

IIIWhich $C_1, C_2$ Make $y$ Real?

Factoring out the real $e^{at}$ leaves $C_1 e^{ibt} + C_2 e^{-ibt}$ , which is real for every $t$ exactly when $C_2 = \overline{C_1}$ .

θ = bt 0.40

💡 Challenge: place $C_2$ so that $y$ is purely imaginary for every $t$ .

IVComplex Form and Real Form

For $y'' + 2y' + 5y = 0$ , the complex-form constants $c, d$ (the real and imaginary parts of $C_1$ ) and the real constants $c_1 = 2c$ , $c_2 = -2d$ describe the same solution.

c 1 d 1 c₁ 2 c₂ −2

With $C_1 = c + id$ and $C_2 = \overline{C_1} = c - id$ :

y = (c + id)\,e^{(-1+2i)t} + (c - id)\,e^{(-1-2i)t}

= e^{-t}\big[c\,(e^{2it} + e^{-2it}) + id\,(e^{2it} - e^{-2it})\big]

= e^{-t}\big[c \cdot 2\cos 2t + id \cdot 2i\sin 2t\big]

= e^{-t}(2c\cos 2t - 2d\sin 2t)

since $i \cdot i = -1$ . Matching $y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$ gives $c_1 = 2c$ and $c_2 = -2d$ .

VQuiz Questions

Problem 1 · The Conjugate Root's Parts

Given: $y'' + 4y' + 13y = 0$ has roots $r = -2 \pm 3i$ . Using the root $r = -2 - 3i$ , find the real and imaginary parts of $e^{rt}$ .

✅ Correct! The real part is

y_1

and the imaginary part is

-y_2

: the same two solutions as from

-2 + 3i

, with one sign flipped.

❌ Not quite. Sine is odd:

\sin(-3t) = -\sin 3t

, so the imaginary part carries a minus sign. These are the parts for the other root,

-2 + 3i

❌ Not quite. Cosine is even:

\cos(-3t) = \cos 3t

, so the real part keeps its plus sign. Only the sine changes sign.

❌ Not quite. Conjugating changes only the sign of the imaginary part of the root. The real part

a = -2

stays, so the factor is

e^{-2t}

, not

e^{2t}

Show solution

By Euler's formula, with $\cos(-3t) = \cos 3t$ (even) and $\sin(-3t) = -\sin 3t$ (odd):

e^{(-2-3i)t} = e^{-2t}\big(\cos(-3t) + i\sin(-3t)\big) = e^{-2t}\cos 3t - i\,e^{-2t}\sin 3t.

Real part $e^{-2t}\cos 3t = y_1$ , imaginary part $-e^{-2t}\sin 3t = -y_2$ . Since $c_1 y_1 + c_2(-y_2) = c_1 y_1 + (-c_2)y_2$ , the conjugate root gives the same general solution.

Problem 2 · Making the Complex Form Real

Given: $y = C_1 e^{(-2+3i)t} + C_2 e^{(-2-3i)t}$ with $C_1 = 3 - 2i$ — find the $C_2$ that makes $y$ real for every $t$ .

✅ Correct!

C_2 = \overline{C_1} = 3 + 2i

, and then

y = e^{-2t}(6\cos 3t + 4\sin 3t)

❌ Not quite.

C_2 = C_1

makes

y

real only when

C_1

is itself real. Here it leaves the imaginary part

-4e^{-2t}\cos 3t

❌ Not quite.

C_2 = -C_1

gives

y = (4 + 6i)\,e^{-2t}\sin 3t

, which is not real. Conjugate

C_1

instead of negating it.

❌ Not quite.

C_2 = -\overline{C_1}

makes the real part vanish:

y

is purely imaginary for every

t

. Drop the minus sign.

Show solution

Replacing $i$ by $-i$ turns $y$ into $\overline{y} = \overline{C_1}\,e^{(-2-3i)t} + \overline{C_2}\,e^{(-2+3i)t}$ . For $y$ to be real, $\overline{y} = y$ ; matching the coefficients of $e^{(-2-3i)t}$ gives

C_2 = \overline{C_1} = \overline{3 - 2i} = 3 + 2i.

Check: $y = 2\,\mathrm{Re}\big[(3 - 2i)\,e^{(-2+3i)t}\big] = e^{-2t}(6\cos 3t + 4\sin 3t)$ , which is real.

Problem 3 · Complex Form to Real Form

Given: $y = (2 - 3i)\,e^{(-1+4i)t} + (2 + 3i)\,e^{(-1-4i)t}$ — write $y$ in real form.

✅ Correct! With

c = 2

d = -3

c_1 = 2c = 4

and

c_2 = -2d = 6

❌ Not quite. The

d

term is

id \cdot 2i\sin 4t

, and

i \cdot i = -1

, so it equals

-2d\sin 4t = 6\sin 4t

. This option treats

i \cdot i

+1

❌ Not quite. The sum and the difference of the exponentials are

2\cos 4t

and

2i\sin 4t

; the factor

2

is missing here.

❌ Not quite. An

i

left in the answer signals an error:

e^{4it} - e^{-4it} = 2i\sin 4t

, not

2\sin 4t

. With the

i

kept,

id \cdot 2i

is real.

Show solution

Here $C_2 = \overline{C_1}$ , with $C_1 = c + id$ , $c = 2$ , $d = -3$ . Factor out $e^{-t}$ and group by $c$ and $d$ :

y = e^{-t}\big[2\,(e^{4it} + e^{-4it}) - 3i\,(e^{4it} - e^{-4it})\big].

By Euler read backwards, $e^{4it} + e^{-4it} = 2\cos 4t$ and $e^{4it} - e^{-4it} = 2i\sin 4t$ :

y = e^{-t}\big[4\cos 4t - 3i \cdot 2i\sin 4t\big] = e^{-t}(4\cos 4t + 6\sin 4t).

Problem 4 · Real Form to Complex Form

Given: $y = e^{3t}(6\cos 2t - 4\sin 2t)$ — find $C_1$ such that $y = C_1 e^{(3+2i)t} + \overline{C_1}\,e^{(3-2i)t}$ .

✅ Correct!

c = c_1/2 = 3

and

d = -c_2/2 = 2

, so

C_1 = 3 + 2i

❌ Not quite. The halving is right, but

c_2 = -2d

, so

d = -c_2/2 = -(-4)/2 = 2

, not

-2

❌ Not quite. The two terms together contribute

2c\cos 2t

, so

c

is half of

c_1

c = 3

, and likewise

d = 2

❌ Not quite.

C_1 = c_1 + ic_2

skips both steps:

c

and

d

are

c_1/2

and

-c_2/2

, not

c_1

and

c_2

Show solution

With $C_1 = c + id$ the complex form equals $e^{3t}(2c\cos 2t - 2d\sin 2t)$ . Match it with $c_1 = 6$ , $c_2 = -4$ :

2c = 6 \Rightarrow c = 3, \qquad -2d = -4 \Rightarrow d = 2.

So $C_1 = 3 + 2i$ . Check: $2\,\mathrm{Re}\big[(3 + 2i)(\cos 2t + i\sin 2t)\big] = 2(3\cos 2t - 2\sin 2t) = 6\cos 2t - 4\sin 2t$ .

Solved: 0 / 4

One Complex Root Is Enough: The Conjugate Pair and the Complex Form of Real Solutions

IKey Formulas

IIThe Conjugate Root Adds Nothing New

IIIWhich C1,C2C_1, C_2C1​,C2​ Make yyy Real?

IVComplex Form and Real Form

VQuiz Questions

IIIWhich $C_1, C_2$ Make $y$ Real?