Ok, your "q" will have an imaginary component, so you cannot use fzero directly. You can, however, solve it using fsolve, where you treat q as a 2-element vector containing the real and imaginary parts.
k = 4;
f = @(q) 1i.*[q(1) + 1i*q(2)]-k.*sin([q(1) + 1i*q(2)]);
x0 = [2; 0.5]
fsolve(@(q) norm(f(q)), x0)
This finds a zero at roughly [3; -0.7]. But since there are multiple minima, the result will depend on the initial conditions.
Note that you can make a plot of f, to visually see roughly where the zeros are. This will help you choose initial conditions.
k = 4;
f = @(q1,q2) 1i.*[q1 + 1i*q2]-k.*sin([q1 + 1i*q2]);
[Q1,Q2] = ndgrid(-10:.2:10);
surf(Q1,Q2, log(abs(f(Q1,Q2))));
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