Simplifying a complex function in order to separate real and imaginary part?

How to Simplify a complex function in order to separate its real and imaginary parts?

For instance, I have the symbolic function 's' as follows and I want to know what is its real part....

syms r theta real
s=(exp(-theta*1i)*((1 - r*exp(theta*1i))^(3/2) - 1))/(3*r);
% real(s)=? imag(s)=?
s1=real(s); % Does not yield the answer. So I try the following command
S=rewrite(s,'sincos');
S1=simplify(collect(real(S))); % Again does not yield the answer....
%s1=real((exp(-theta*1i)*((1 - r*exp(theta*1i))^(3/2) - 1))/r)/3
%S1=real((((1 - r*(cos(theta) + sin(theta)*1i))^(3/2) - 1)*(cos(theta) - sin(theta)*1i))/r)/3

[KSSV]

Why do you think real(s) is not givng you the realpart of s?

[zah ra]

I want some symbolic function in 'r' and 'theta' variables as the real part of 's' (independent from the basic imaginary unit 'i')

Answer 1

[Ameer Hamza]

Due to the term (1 - r*exp(theta*1i))^(3/2) in your expression, it might not be possible to write an explicit expression (maybe an infinite series is possible) for the real and complex parts. However, if you just care about the numeric output, the real(s) does give real values

syms r theta

s = (exp(-theta*1i)*((1 - r*exp(theta*1i))^(3/2) - 1))/(3*r);
S = real(s);
S = matlabFunction(S, 'Vars', {'r', 'theta'});
v = S(2, 0:0.1:10); % some values
tf = all(v==real(v)); % test if all are real
tf

Result

>> test
tf =
  logical
   1

[zah ra]

Thanks for your answer but I want some function in 'r' and 'theta' variables as the real part of 's' (independent from the basic imaginary unit 'i') , a function such as

(cos(θ) cos(3/2 arg(1 - e^(i θ) r)) (r^2 sin^2(θ) + (1 - r cos(θ))^2)^(3/4))/(3 r) + (sin(θ) sin(3/2 arg(1 - e^(i θ) r)) (r^2 sin^2(θ) + (1 - r cos(θ))^2)^(3/4))/(3 r) - cos(θ)/(3 r)

However, arg(1 - e^(i θ) r) is still needed to be simplified....

[Ameer Hamza]

Yes, I mentioned that because of the term (1 - r*exp(theta*1i))^(3/2), such an explicit representation of real and imaginary parts does not seem to be possible. If this term had an integer exponent, it could be easily expanded into a finite number of terms and separated into real and imaginary parts. But for fractional powers, such expansion is not possible.

[zah ra]

O.K. thanks so much ...

[Ameer Hamza]

I am glad to be of help!

Answer 2

[David Goodmanson]

Hi zah ra,

An explicit form for real and imaginary parts of

(1 - r*exp(theta*1i))^(3/2)

is possible, although not particularly convenient. You are using the 3/2 power, but the code below works for any (real) power n. Also the code is for r<1, but with a bit more work it can be modified for the case r>1.

r = .6;
th = 0:.001:2*pi;
n = 1.536;

% exact expressions, want to reproduce these
re1 = real((1-r*exp(i*th)).^n);
im1 = imag((1-r*exp(i*th)).^n);            

% obtain equivalent expressions in terms of real functions
a = sqrt(r^2+1-2*r*cos(th));    % third side of the 1(one),r,a triangle
phi = -asin((r./a).*sin(th));   % angle between 'a' and the x axis
re2 = a.^n.*cos(n*phi);
im2 = a.^n.*sin(n*phi);

% compare with re1,im1.
% re2 and im2 are offset for visual purposes
plot(th,[re1;im1;re2+.05;im2+.05])

After finding the expressions for real and imag, you can go back to symbolic multiplication to obtain the real and imaginary parts of s. But as is usually the case, It's a lot of trouble to recreate complex algebra in terms of real quantities, and the resulting jumble of code is not particularly revealing. Maybe there is good reason to do that in your case. But most of the time, numerical results are ultimately what is wanted. Then it is a lot more convenient, compact and error-free to allow complex algebra to do its thing and use real(s) and imag(s) afterward.

[zah ra]

Thank you very much for your attention and guidance ...