How to separate real and imaginary components from a given expression in Matlab?

Hello everyone I have a expression and I need to separate the real and imaginary components of the expression. I have written a simple code using symbolic math toolbox. The code is:

clc
syms X1 X2 XC RL

Zin=((X1*XC)-1i*(X2*RL))/(RL+1i*(X1*X2*XC));
A=real(Zin)
B=imag(Zin)

Can anyone please point out whats wrong with it or maybe another way to do this right.

Thank You.

Answer 1

[Steven Lord]

syms X1 X2 XC RL

Zin=((X1*XC)-1i*(X2*RL))/(RL+1i*(X1*X2*XC));
A=real(Zin)

A = 

B=imag(Zin)

B = 

What leads you to believe something is wrong? For general X1, X2, XC, and RL (which could be complex) that's the correct result. If you were to assume that they were real you get something a bit different.

syms X1 X2 XC RL real
A = real(Zin)

A = 

B = imag(Zin)

B = 

[Md. Golam Zakaria]

@Steven Lord, I suspected wrong because of the i sign. There should not be any 'i' operator in the separated forms , right?

[Torsten]

By default, all symbolic variables are assumed complex-valued. But you might assume that some or all of them are real-valued:

syms z1 z2
z = z1 + 1i*z2;
real(z)

ans = 

imag(z)

ans = 

syms z1 z2 real
z = z1 + 1i*z2;
real(z)

ans = 

imag(z)

ans = 

Do you understand the difference ?

[Md. Golam Zakaria]

@Steven Lord, I do. Thanks.

[Walter Roberson]

@Md. Golam Zakaria

We are not told that X1 X2 XC RL are real valued.

Suppose we had the expression 1i * X1 * X2 and we do not know that X1 and X2 are real. If we take real(1i * X1 * X2) is it correct that would be the same as X1*X2 ? Obviously not: real(1i * 3 * 5) = real(15i) = 0, which is not the same as real(3*5) = real(15) = 15. Should we guess then that real(1i * X1 * X2) = real(conj(X1 * X2)) ?

format long g
A = (3+4i) * (5+6i)

A = 
                         -9 +                    38i

real(1i * A)

ans = 
   -38

conj(A)

ans = 
                         -9 -                    38i

real(conj(A))

ans = 
    -9

real(conj(3+4i) * conj(5+6i))

ans = 
    -9

imag(3+4i) * imag(5+6i)

ans = 
    24

No. However it is correct that real(1i*A) == imag(A). Whether you write imag(A) or real(1i*A) is a matter of taste.

[Steven Lord]

What's the real part of x+1i*y? If you said x you'd be incorrect in the general case, as shown by this example that substitutes numbers for x and y:

x = 1+2i;
y = 3+4i;
z = x + 1i*y

z = -3.0000 + 5.0000i

RE = real(z)

RE = -3

IM = imag(z)

IM = 5

RE == x % false

ans = logical
   0