% Subtraction can be defined in terms of addition
% and the additive inverse.
%
% If
%
%     w = a + bj
%
% and
%
%     z = c + dj,
%
% then
%
%     w - z = w + (-z)
%
%           = (a + bj) + (-c - dj)
%
%           = (a - c) + (b - d)j.
%
% Thus, complex numbers are subtracted by
% subtracting their real components and their
% imaginary components separately.

w = 5.3 + 3.2j
z = 3.2 - 4.7j

fprintf('\nCalculations with subtraction:\n')

w - z

% We can obtain the same result by adding the
% additive inverse of z to w.

w + (-z)

% Addition is commutative, so we can also add
% the two numbers in the opposite order.

(-z) + w

% Thus,
%
%     w - z = w + (-z) = (-z) + w.

fprintf('\nEquivalent ways of calculating w - z:\n')

w - z
w + (-z)
(-z) + w

% We can also see how subtraction is performed
% by separately subtracting the real and
% imaginary components.

real(w) - real(z)
imag(w) - imag(z)

fprintf('Re(w) - Re(z) = %f\n', ...
        real(w) - real(z))

fprintf('Im(w) - Im(z) = %f\n', ...
        imag(w) - imag(z))

% Putting these two components back together
% gives the difference of the two complex
% numbers.

fprintf('(Re(w) - Re(z)) + j(Im(w) - Im(z)) = %f%+fj\n\n', ...
        real(w) - real(z), ...
        imag(w) - imag(z))

% Thus,
%
%     Re(w - z) = Re(w) - Re(z)
%
% and
%
%     Im(w - z) = Im(w) - Im(z).


% Unlike addition, subtraction is not
% commutative. In general,
%
%     w - z ~= z - w.
%
% Instead, reversing the order gives the
% additive inverse:
%
%     w - z = -(z - w).

fprintf('\nReversing the order of subtraction:\n')

w - z
z - w
-(z - w)

% Confirm that
%
%     w - z = -(z - w).

(w - z) == -(z - w)


% Subtracting a number from itself gives zero.

fprintf('\nSubtracting a number from itself:\n')

w - w
z - z

% Subtracting zero does not change a number.

fprintf('\nSubtracting zero:\n')

w - 0
z - 0

% Subtracting a number from zero gives its
% additive inverse.

fprintf('\nSubtracting a number from zero:\n')

0 - w
-w

0 - z
-z

% Thus,
%
%     w - 0 = w
%
% while
%
%     0 - w = -w.


% Subtracting an additive inverse is equivalent
% to addition:
%
%     w - (-z) = w + z.

fprintf('\nSubtracting an additive inverse:\n')

w - (-z)
w + z

(w - (-z)) == (w + z)


% Parentheses are important when more than two
% numbers are being subtracted. Subtraction is
% not associative. In general,
%
%     (w - z) - v ~= w - (z - v).

v = -1.8 + 2.1j

fprintf('\nSubtraction is not associative:\n')

(w - z) - v
w - (z - v)

% The first expression can be written as
%
%     (w - z) - v
%       = w + (-z) + (-v)
%       = w - z - v.
%
% The second expression is
%
%     w - (z - v)
%       = w - z + v.
%
% Consequently, these expressions generally
% have different values.

((w - z) - v) == (w - (z - v))