HP 15c - Manual (Page 70)

70 Section 3: Calculating in Complex Mode
70
negative values of Re(z). This would slow convergence considerably unless the first guess z
0
were extremely close to a root. In addition, this f(z) vanishes infinitely often, so it's difficult
to determine when all desired roots have been calculated. But by rewriting this equation as
e
z
= −8/(z + 9)
and taking logarithms, you obtain an equivalent equation
z = ln(−8/(z + 9)) ± i2nπ for n = 0, 1, 2, ….
This equation has only two complex conjugate roots z for each integer n. Therefore use the
equivalent function
f(z) = z − ln(−8/(z + 9)) ± i2nπ for n = 0, 1, 2, ….
and apply Newton’s iteration
z
k + 1
= z
k
− (z
k
− ln(−8/(z
k
+ 9)) ± i2nπ) / (1 + 1/(z
k
+ 9)).
As a first guess, choose z
0
as A(n), the approximation given earlier. A bit of algebraic
rearrangement using the fact that ln(±i) = ±i π/2 leads to this formula:
z
k + 1
= A(n) + ((z
k
− A(n)) + (z
k
+ 9)ln(iIm(A(n)) / (z
k
+ 9))) / (z
k
+ 10).
In the program below, Re(A(n)) is stored in R
0
and Im(A(n)) is stored in R
1
. Note that only
one of each conjugate pair of roots is calculated for each n.
Keystrokes Display
|¥
Program mode
´ CLEARM 000-
´bA 001-42,21,11 Program for A(n).
|"8 002-43, 5, 8 Specifies real arithmetic.
v 003- 36
+ 004- 40
. 005- 48
5 006- 5
+ 007- 40
|$ 008- 43 26
* 009- 20 Calculates (2n + ½)π.
v 010- 36
O1 011- 44 1
8 012- 8
÷ 013- 10
|N 014- 43 12
” 015- 16 Calculates –ln((2n + ½)π/8).