$z\tan z-a$ has only real roots for $a>0$

Let \[ z=x+iy \] be a zero of \(z\tan z-a\), so \[ z\tan z=a>0. \] Using \[ \tan(x+iy) =\frac{\sin 2x+i\sinh 2y} {\cos 2x+\cosh 2y}, \] we obtain \[ z\tan z =\frac{x\sin 2x-y\sinh 2y +i\bigl(x\sinh 2y+y\sin 2x\bigr)} {\cos 2x+\cosh 2y}. \]

Suppose \(y\ne0\). Then \[ \cos 2x+\cosh 2y>0, \] and since \(z\tan z\) is real, its imaginary part vanishes: \[ x\sinh 2y+y\sin 2x=0. \] Thus \[ \sin 2x=-\frac{x}{y}\sinh 2y. \] Substituting this into the real part gives \[ \operatorname{Re}(z\tan z) =-\frac{(x^2+y^2)\sinh 2y} {y(\cos 2x+\cosh 2y)}. \] For \(y\ne0\), \[ \frac{\sinh 2y}{y}>0, \] so the right-hand side is strictly negative. This contradicts \[ z\tan z=a>0. \]

Therefore \(y=0\), and every zero is real.