Lecture Notes in Differential Equations - Bruce E. Shapiro

157
For k = 0, 1, 2, . . . , n − 1 the right hand side produces unique results. But
for k ≥ n, the results start to repeat: k = n gives the same angle as k = 0;
k = n + 1 gives the same angle as k = 1; and so on. Hence there are
precisely n unique numbers.
Example 18.3. Find the three cube roots of 27.
To find the cube roots we repeat the proof!
27 = 27 + (0)i = 27e (i)(0) = 27e (i)(0+2π) = 27e 2kπi (18.38)
3√
27 = 27 1/3 (e 2kπi ) 1/3 (18.39)
= 3e 2kπi/3 (18.40)
For k = 0 this gives
3√
27 = 3 (18.41)
For k = 1 this gives
3√
27 = 3e 2πi/3 = 3
For k = 2 this gives
3√
27 = 3e 4πi/3 = 3
(
cos 2π 3 + i sin 2π )
= − 3 3 2 + i3√ 3
2
(
cos 4π 3 + i sin 4π )
= − 3 3 2 − i3√ 3
2
(18.42)
(18.43)
Using k = 3 will give us the first result, and so forth, so these are all the
possible answers.
Theorem 18.6. If the roots of
ar 2 + br + c = 0 (18.44)
are a complex conjugate pair
}
r 1 = µ + iω
r 2 = µ − iω
(18.45)
where µ, ω ∈ R and ω ≠ 0, (this will occur when b 2 < 4ac), then the solution
of the homogeneous second order linear ordinary differential equation with
constant coefficients
Ly = ay ′′ + by ′ + cy = 0 (18.46)
is given by
y H = C 1 e r1t + C 2 e r2t (18.47)
= e µt (A cos ωt + B sin ωt) (18.48)