求方程：z²+5+12i=0的解.
z²=-5-12i=13{cos[π+arctan(12/5)]+isin[π+arctan(12/5)]}
=(12/5)[-cosarctan(12/5)-isinarctan(12/5)]
故z=-[(2/5)√15]{cos[2kπ+arctan(12/5)]/2+isin[2kπ+arctan(12/5)]/2},(k=0,1)
即z₁=-[(2/5)√15]{cosarctan(12/5)+isinarctan(12/5)]=-[(2/5)√15][(5/13)+(12/13)i]
=-(2√15)/13-i(24√15)/65]
z₂=-[(2/5)√15]{cosarctan(12/5)-isinarctan(12/5)]=-[(2/5)√15][(5/13)-(12/13)i]
=-(2√15)/13+i(24√15)/65]