上 5−2(z 3)=z−4 292543-2/3z-4+z=5+2/3(z-6)
Figure 3 Applying (4) to z 1 = z 2 = −44i = 4 √ 2e3 4 πi (our earlier example), we get (−44i)2 = (4 √ 2e34πi)2 = 32e 3 2 πi = −32i By an easy induction argument, the formula in (4) can be used to prove that for any positive integer n If z = reiθ, then zn = rneinθ This makes it easy to solve equations like z3 = 1 Indeed, writing the unknown number z as reiθ, we have r3ei3θ