Thread Subject: Real and imaginary parts of symbolic expression

Dear All,

I experience a problem getting the real and imaginary parts
of a complex expression in symbolic form.
For example I have this code and will like to know how to
get the real and imaginary parts:

clear; clc;
syms x y n
z=n*cos(x)+i*y*sin(x);
r = real(z);
m = imag(z);

The expressions I expected to get were:
r = n*cos(x) and
m = y*sin(x).

But instead I get:
r =
 
1/2*n*cos(x)+1/2*i*y*sin(x)+1/2*conj(n*cos(x)+i*y*sin(x))
 
 and

m =
 
-1/2*i*(n*cos(x)+i*y*sin(x)-conj(n*cos(x)+i*y*sin(x)))

Though these are correct, I cannot they are more
complicated and I cannot simplify them to obtain the
answers I listed above. Can anyone help me out of this?
I need to have a way of obtaining the real and imaginary
parts of any arbitrary complex expression is the actual
form without having longer expressions.

Thanks

ES

In article <g4kl0p$k1c$1@fred.mathworks.com>,
Emeka Obe <obeway@gmail.com> wrote:

>I experience a problem getting the real and imaginary parts
>of a complex expression in symbolic form.
>For example I have this code and will like to know how to
>get the real and imaginary parts:

>clear; clc;
>syms x y n
>z=n*cos(x)+i*y*sin(x);
>r = real(z);

>The expressions I expected to get were:
>r = n*cos(x) and

>But instead I get:
>r =

>1/2*n*cos(x)+1/2*i*y*sin(x)+1/2*conj(n*cos(x)+i*y*sin(x))

The output you received is correct, and your expected output is not.

Your variable y is symbolic, so the program is to produce the general
output for all possible values the symbol y could assume. That
output is for real() is *not* n*cos(x) . Simple proof: suppose y was 5*i .
real(n*cos(x) + i*i*5*sin(x)) is real(n*cos(x) - 5*sin(x))
which is different than your proposed answer of n*cos(x) .

The hidden assumption in your proposed expression is that none of
n, x, or y are complex: if any of them are complex then your
proposed output is incorrect. If that assumption is in fact valid
for your problem domain, then in order to have your proposed output
appear automatically, tell matlab about those assumptions.

I do not have the symbolic toolbox, so I do not know what the
syntax is for providing assumptions. In the base Maple engine
which underlies the symbolic toolbox, the way I would code would be
(in Maple code!):

z := n*cos(x) + I*y*sin(x);
r := Re(z) assuming real;
m := Im(z) assuming real;


--
  "There is no greater calling than to serve your fellow men.
   There is no greater contribution than to help the weak.
   There is no greater satisfaction than to have done it well."
                                              -- Walter Reuther

"Walter Roberson" <roberson@ibd.nrc-cnrc.gc.ca> wrote in message
news:g4lm76$q9h$1@canopus.cc.umanitoba.ca...
> In article <g4kl0p$k1c$1@fred.mathworks.com>,
> Emeka Obe <obeway@gmail.com> wrote:

*snip*

> I do not have the symbolic toolbox, so I do not know what the
> syntax is for providing assumptions. In the base Maple engine
> which underlies the symbolic toolbox, the way I would code would be
> (in Maple code!):

The syntax for assuming that x, y, and n are real symbolic variables is:

syms x y n real

--
Steve Lord
slord@mathworks.com