For which values $k \in \mathbb{R}$ does $kz - \tan z$ have non-real roots? This is not a homework or university course question, it is one purely of my own posing that arose while investigating the Weierstraß factorization of various complex functions.
Numerical Examples suggests that this is the case for $0<k<1$, however, it is not immediately obvious how one would prove such a result without resorting to tedious multivariable calculus.
 A: (See Real and imaginary part of $\tan(a+bi)$ )
$$\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2},\quad\tan(x+iy)=\frac{\sin2x+i\sinh2y}{\cos2x+\cosh2y}$$
$$k=\frac{\tan(x+iy)}{x+iy}=\frac{(x\sin2x+y\sinh2y)+i(x\sinh2y-y\sin2x)}{(x^2+y^2)(\cos2x+\cosh2y)}$$
This shows that $\tan(x+iy)/(x+iy)$ is real exactly when
$$x\sinh2y=y\sin2x.$$
We want to know whether there are solutions with $y\neq0$. If also $x\neq0$, then we have
$$\frac{\sinh2y}{2y}=\frac{\sin2x}{2x};$$
but the left side is always greater than $1$, while the right side is always less than $1$. So we must take $x=0$.
$$k=\frac{\tan(iy)}{iy}=\frac{\tanh y}{y}$$
Solutions come in conjugate pairs: $\tanh(-y)/(-y)=\tanh(y)/y$, so we can assume $y>0$. This function is strictly positive, and less than $1$, and has limiting values
$$\lim_{y\to0}\frac{\tanh y}{y}=1,\qquad\lim_{y\to\infty}\frac{\tanh y}{y}=0,$$
and has derivative
$$\frac{\text{sech}^2y}{y}-\frac{\tanh y}{y^2}=\frac{2y-\sinh2y}{2y^2\cosh^2y}<0.$$
Therefore, for any $0<k<1$, there is exactly one $y>0$ (thus exactly two $y\neq0$) such that $\tan(iy)/(iy)=k$; and for any other value of $k\in\mathbb R$, there is no such $y$.
