More fun in 2016, Part 12

should not take more than a second if you are familiar with this. easy proof is that while $z$ is a root, so is $z\pm2\pi$ . so we look at the primary solution $z = i\ln(2016\pm \sqrt{4064255})$ , then all other roots are $z = i\ln(2016\pm \sqrt{4064255}),i\ln(2016\pm \sqrt{4064255})\pm2\pi,i\ln(2016\pm \sqrt{4064255})\pm4\pi.......$ hence infinite solutions.