已知虚数z满足|z|=√2，且(z-a)=a^2,求实数a.

z = 2^(1/2)[cosb + isinb], 0 <= b < 2PI.

a^2 = z - a = 2^(1/2)[cosb + isinb] - a = 2^(1/2)cosb - a + i2^(1/2)sinb,

2^(1/2)sinb = 0, b = 0或PI.
cosb = 1或-1.
a^2 = 2^(1/2)cosb - a,
a^2 + a - 2^(1/2)cosb = 0,

当b = PI,cosb = -1时，
1 + 4*2^(1/2)cosb < 0,方程a^2 + a - 2^(1/2)cosb = 0关于a没有实数解,不符题意。

因此，b = 0. cosb = 1.
0 = a^2 + a - 2^(1/2)cosb = a^2 + a - 2^(1/2),

a = {-1 + [1 + 4*2^(1/2)]^(1/2)}/2
或
a = {-1 - [1 + 4*2^(1/2)]^(1/2)}/2