[R] finding complex roots in R

[Ravi Varadhan]

Hi Toros,

It is actually quite easy to do this with existing real-root finders.  If
you want to solve for F(z) = 0, where z = x + iy, you simultaneously solve
for the zeros of the real and imaginary part of F.  

Here are a couple of examples using the package "BB":

fn1 <- function(x, a){
z <- x[1] + 1i * x[2]
f <- exp(z) + a
c(Re(f), Im(f))
}

fn2 <- function(x, a){
z <- x[1] + 1i * x[2]
f <- sin(z)^2 + sqrt(z) - log(z)
c(Re(f), Im(f))
}

require(BB)

# First example
> BBsolve(par=c(1,1), fn=fn1, a=1)
  Successful convergence.
$par
[1] -4.799024e-10  3.141593e+00

$residual
[1] 4.819472e-10

$fn.reduction
[1] 0.0002550771

$feval
[1] 81

$iter
[1] 5

$convergence
[1] 0

$message
[1] "Successful convergence"

$cpar
method      M     NM 
     2     50      1 

So, the complex root is: 0 + i*pi, as you might have guessed.

# Second example
> BBsolve(par=c(1,1), fn=fn2)
  Successful convergence.
$par
[1] 0.2555197 0.8948303

$residual
[1] 9.079212e-08

$fn.reduction
[1] 0.0002494436

$feval
[1] 168

$iter
[1] 92

$convergence
[1] 0

$message
[1] "Successful convergence"

$cpar
method      M     NM 
     2     50      1 

Here the complex root is: 0.2555 + i*0.8948.

In these two examples the function F(z) is scalar.  We can readily extend
this to vector F using the method that I just described.

Hope this helps,
Ravi.

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of Toros Caglar
Sent: Wednesday, September 15, 2010 11:33 PM
To: r-help at r-project.org
Subject: [R] finding complex roots in R

I am looking for a way to find the roots of a non-polynomial
expression. I know R has a few ways to deal with polynomials, but, I
could not find a method that deals with functions involving e^(x) type
arguments, that have complex roots as well as real roots.

Any ideas?

Thanks,

Toros

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