HP 15c - Manual (Page 69)

Section 3: Calculating in Complex Mode 69
Keystrokes Display
¦ 0.9980 Calculates z
1
(real part).
´% (hold) 0.0628 Imaginary part of z
1
.
50 O V 50.0000 Stores root number in Index
register.
¦ -1.0000 Calculates z
50
(real part).
´% (hold) 0.0000 Imaginary part of z
50
Solving an Equation for Its Complex Roots
A common method for solving the complex equation f(z) = 0 numerically is Newton's
iteration. This method starts with an approximation z
0
to a root and repeatedly calculates
z
k + 1
= z
k
– f(z
k
) / f’(z
k
)
Until z
k
converges.
The following example shows how _
can be used with Newton’s iteration to estimate
complex roots. (A different technique that doesn't use Complex mode is mentioned on page
18.)
Example: The response of an automatically controlled system to small transient
perturbations has been modeled by the differential delay equation
0 ) 1 ( 8 ) ( 9 ) (
   
t w t w t w
dt
d
.
How stable is this system? In other words, how rapidly do solutions of this equation decay?
Every solution w(t) is known to be expressible as a sum


k
zt
e z c t w ) ( ) (
involving constant coefficients c(z) chosen for each root z of the differential-delay equation's
associated characteristic equation:
z + 9 + 8e
−z
= 0
Every root z = x + iy contributes to w(t) a component e
zt
= e
xt
(cos(yt) + i sin(yt)) whose rate
of decay is faster as x, the real part of z, is more negative. Therefore, the answer to the
question entails the calculation of all the roots z of the characteristic equation. Since that
equation has infinitely many roots, none of them real, the calculation of all roots could be a
large task.
However, the roots z are known to be approximated for large integers n by
z ≈ A(n) = -ln((2n + ½)π/8) ± i(2n + ½) π for n = 0, 1,2, .... The bigger is n, the better is the
approximation. Therefore you need calculate only the few roots not well approximated by
A(n) —the roots with |z| not very big.
When using Newton's iteration, what should f(z) be for this problem? The obvious function
f(z) = z + 9 + 8e
-z
isn't a good choice because the exponential grows rapidly for larger