Break the limit

We need to solve $\sin2z = 14$ . Writing $2z=x+iy$ we deduce that $\sin x \cosh y + i\cos x \sinh y \; = \; 14$ so we need $\cos x = 0, \sin x = 1$ and $\cosh y = 14$ . Thus $x = (2n+\tfrac12)\pi$ for integer $n$ , and $y = \mp\ln(14+\sqrt{195}) = -\ln(14\pm\sqrt{195})$ . Thus $z \; = \; \tfrac12(x+iy) \; =\; n\pi + \tfrac14\pi - \tfrac12i\ln(15\pm\sqrt{195})$ for integer $n$ , making the answer $4+2+14+195 = \boxed{215}$ .

Rewriting $sin(z)$ and $cos(z)$ we get

$\dfrac{e^{iz}-e^{-iz}}{2i}\cdot\dfrac{e^{iz}+e^{-iz}}{2}=7 \iff \dfrac{e^{2iz}-e^{-2iz}}{4i}=7$

Multiply both sides of the equation by $4ie^{2iz}$ to get a quadratic in $e^{2iz}$

$(e^{2iz})^2-28ie^{2iz}-1=0$

Solving for $e^{2iz}$

$e^{2iz}=14i\pm\sqrt{(14i)^2+1}=14i\pm\sqrt{-195}=i(14\pm\sqrt{195})$

Taking the natural logarithm of both sides and using the fact that $log(ab)=log(a)+log(b)$

$2iz=ln[i(14\pm\sqrt{195})]=ln(i)+ln(14\pm\sqrt{195})$

To calculate $ln(i)$ we express $i$ in its polar form

$ln(i)=ln(e^{i\frac{\pi}{2}+2\pi ni})=i\left(\frac{\pi}{2}+2\pi n\right) \qquad n\in\mathbb{Z}$

Substituting back and solzing for $z$

$z=\dfrac{1}{2i}\left[i\left(\frac{\pi}{2}+2\pi n\right)+ln(14\pm\sqrt{195})\right]=\dfrac{\pi}{4}-\dfrac{i}{2}ln(14\pm\sqrt{195})+\pi n$

Therefore the solution is

$4+2+14+195=\boxed{215}$