By my understanding of the question, Vafa has it pretty well covered. However I think it only wants the one solution where arg(z) is a minimum positive. By pythagoras as you can see in Vafa's diagram, the modulus here will be 4. I found it more useful to express the argument in a cleaner form, like tan inverse (4/3), seeing as arguments are usually put in this form. so my answer was z = 4(cos(tan inverse (4/3)) + isin(tan inverse (4/3)) ). Using a triangle you can express it as z=4(cos(cos inverse(3/5) + isin(sin inverse(4/5)) ie z= 12/5 + 16i/5