complex number!!!! 求救!!!!

|z - 3| + |z + 3| = 10

情況一：當 z ≤ -3
|z - 3| = -(z - 3) 及 |z + 3| = -(z + 3)
所以 -(z - 3)- (z + 3) = 10 及 z ≤ 3
-z + 3 - z - 3= 10 及 z ≤ 3
-2z = 10 及 z ≤ 3
z = -5 及 z ≤ -3
故情況一的解為 z = -5

情況二：當 -3 ≤ z ≤ 3
|z - 3| = -(z - 3) 及 |z + 3| = z + 3
所以 -(z - 3) +(z + 3) = 10 及 -3 ≤ z ≤ 3
-z + 3 + z + 3= 10 及 -3 ≤ z ≤ 3
6 = 10 及 -3 ≤ z ≤ 3
故情況二無解。

情況三：當 z ≥ 3
|z - 3| = z - 3 及 |z + 3| = z + 3
所以 (z - 3) +(z + 3) = 10 及 z ≥ 3
z - 3 + z + 3= 10 及 z ≥ 3
2z = 10 及 z ≥ 3
z = 5 及 z ≥ -3
故情況三的解為 z = 5

綜合以上三個情況，此題的解為：
z = -5 或 z = 5

2013-07-20 19:19:01 補充：
|z - 3| + |z + 3| = 10

Case I：When z ≤ -3
|z - 3| = -(z - 3) and |z + 3| = -(z + 3)
Hence, -(z - 3)- (z + 3) = 10 and z ≤ 3
-z + 3 - z - 3= 10 and z ≤ 3
-2z = 10 and z ≤ 3
z = -5 and z ≤ -3
Hence, the solution for Case I is z = -5

2013-07-20 19:19:18 補充：
Case II：When -3 ≤ z ≤ 3
|z - 3| = -(z - 3) and |z + 3| = z + 3
Hence, -(z - 3) +(z + 3) = 10 and -3 ≤ z ≤ 3
-z + 3 + z + 3= 10 and -3 ≤ z ≤ 3
6 = 10 and -3 ≤ z ≤ 3
Hence, Case II has no solution.

2013-07-20 19:19:52 補充：
Case III：When z ≥ 3
|z - 3| = z - 3 and |z + 3| = z + 3
Hence, (z - 3) +(z + 3) = 10 and z ≥ 3
z - 3 + z + 3= 10 and z ≥ 3
2z = 10 and z ≥ 3
z = 5 and z ≥ -3
Hence, the solution for Case III is z = 5

The solution of this question is :
z = -5 or z = 5 ...... (ans)

Let complex number z = a+bi, where a,b are real numbers

Then the expression can be rewritten as
|a+bi-3| + |a+bi+3| = 10
√[(a-3)^2+b^2] + √[(a+3)^2+b^2] = 10
(a-3)^2+b^2 + 2√[(a-3)^2+b^2)]√[(a+3)^2+b^2] + (a+3)^2+b^2 = 10^2
a^2-6a+9 +2√[((a-3)^2+b^2)((a+3)^2+b^2)] + a^2+6a+9+b^2 = 100
√[(a-3)^2(a+3)^2 + ((a-3)^2+(a+3)^2)b^2 + b^4] = (82-2a^2-2b^2)/2 = 41-a^2-b^2
(a^2-6a+9)(a^2+6a+9) + (2a^2+18)b^2 + b^4 = 1681 + a^4 + b^4 - 82a^2 - 82b^2 -2a^2b^2
a^4-18a^2+81 + 2a^2b^2 + 18b^2 + b^4 = 1681 + a^4 + b^4 - 82a^2 - 82b^2 -2a^2b^2
64a^2 + 100b^2 = 1600
16a^2 + 25b^2 = 400

Therefore the solutions are z = a+bi,
where a,b are real numbers satisfying 16a^2 + 25b^2 = 400

If z is real, then b=0 and a=±5. i.e z=±5
This is the same as the other solution which assumes z is a real number

If you are not satisfied with the solution or plan to remove the question, please notify me

2013-07-21 10:03:58 補充：
You are only required to square both sides of the equations until all root signs are gone. It is not very difficult.

2013-07-21 22:33:47 補充：
Corrections:
4th line:
a^2-6a+9+b^2+2√[((a-3)^2+b^2)((a+3)^2+b^2)]+a^2+6a+9+b^2 = 100
6th line:
(a^2-6a+9)(a^2+6a+9)+(2a^2+18)b^2+b^4 = 1681+a^4+b^4-82a^2-82b^2+2a^2b^2
7th line:
a^4-18a^2+81+2a^2b^2+18b^2+b^4 = 1681+a^4+b^4-82a^2-82b^2+2a^2b^2