懂答這問題有10分^^ thxx

Let (x + yi)^2 = 10i

(x^2 - y^2) + 2xyi = 10i

So, x^2 - y^2 = 0 and 2xy = 10

x = y or x = -y

When x = y, 2y^2 = 10 => y = √5 or -√5

When x = -y, -2y^2 = 10 => y^2 = -5 => y = √5i or -√5i

So, the square roots are √5 + √5i, √5 - √5i, -√5 + √5i, √5 + √5i


2011-09-17 18:31:40 補充：
So, the square roots are √5 + √5i, -√5 - √5i, -√5 - √5i, -√5 + √5i

2011-09-17 18:39:24 補充：
From line 5

When x = y, 2y^2 = 10 => y = √5 or -√5

When x = -y, -2y^2 = 10 => y^2 = -5 (rejected as y is real)

So, the square roots are √5 + √5i, -√5 - √5i

Let a = √(10i) s.t. a = x+yi for x,y are real numbers.
a^2 = 10i => (x+yi)^2 = 10i
x^2 + 2xyi - y^2 = 10i
Real part: x^2 - y^2 = 0 -----------(1)
Imaginary part: 2xy = 10 => xy=5 ----------------(2)

2011-09-17 19:40:15 補充：
x^2 = y^2 => x = -y or x = y
When x = y, y^2 = 5 => x,y = √5 or -√5
When x = -y, -y^2 = 5(rej.)
Hence, a = √5 + √5i or -√5 - √5i

2011-09-17 19:41:15 補充：
Checking:
(√5+√5i)^2 = 5 + 2(5)i - 5 = 10i .which is correct
(-√5-√5i)^2 = (√5+√5i)^2 , which is also correct

There are only 2 roots, not 4 roots