Plotting $α^2\sin^2 θ-\sin^2(αθ)=0$, where $α≠0$ is an arbitrary complex number Let $$α^2 \sin^2 θ-\sin^2 (αθ)=0,$$
where $α≠0$, and $α=a+ib$, $θ∈(0,2π]$.
I want to draw a graph between $\mathrm{Re}\,α$ and $θ$ and a graph between $\mathrm{Im}\,α$ and $θ$. The real part of $α$ is between the interval $(0, 3)$.
Any help or hint to write the Matlab program. Thanks in advance.
 A: You can use the fact that:
$$\sin(a+ib)=\sin a\cosh b+i\cos a\sinh b$$
So that:
$$\sin^2(a+ib)=\sin^2 a\cosh^2 b - \cos^2 a\sinh^2 b +\frac{i}{2}\sin (2a)\sinh (2b)$$
And:
$$\alpha^2\sin^2\theta−\sin^2(\alpha\theta)=$$
$$(a^2-b^2+2abi)\sin^2\theta−\sin^2(\alpha\theta)=$$
$$(a^2-b^2+2abi)\sin^2\theta−\sin^2 (a\theta)\cosh^2 (b\theta) + \cos^2 (a\theta)\sinh^2 (\theta b) -\frac{i}{2}\sin (2a\theta)\sinh (2b\theta)=0$$
You want to plot $a$ or $b$ vs. $\theta$. to that end, compare the real and imaginary parts to get:
$$2ab\sin^2\theta=\frac{1}{2}\sin (2a\theta)\sinh (2b\theta)$$
$$(a^2-b^2)\sin^2\theta−\sin^2 (a\theta)\cosh^2 (b\theta) + \cos^2 (a\theta)\sinh^2 (\theta b)=0$$
These two equations are rather intractable, so really the only thing you can do is loop over $\theta$, and for each value solve these two equations for $a,b$ numerically. The fact that there is more than one solution per $\theta$ makes this a very tough problem, as numerical solvers that use gradient descent like methods will only converge to one of the many possible solutions.
