解:z1·z2=(a+bi)(3+4i)=(3a-4b)+(4a+3b)i.
∵z1·z2是纯虚数,
∴3a-4b=0且4a+3b≠0, ①
且a2+b2=25. ②
由①和②,得或
∴z1=4+3i或z1=-4-3i.