maths t 问题

薰草

if ( x +iy)^2=i ,find all the real values of x and y .
how to solve tis?

[ 本帖最后由 薰草 于 7-6-2009 11:06 PM 编辑 ]

Wongkokchoy

先expend Left hand side, 然后equals real part wif real part, imaginary wif imaginary part

Wongkokchoy

先expend Left hand side, 然后equals real part wif real part, imaginary wif imaginary part

薰草

哦.....谢谢...

文锋

原帖由 薰草 于 7-6-2009 11:05 PM 发表
if ( x +iy)^2=i ,find all the real values of x and y .
how to solve tis?

已修改!
Given : ( x +iy)^2=i
(x^2 - y^2) + i (2xy) = i
so,x^2 - y^2= 0
  (x-y)(x+y) = 0 ---------- (1)
     2xy = 1 -------------- (2)

From Eqn (2).
y = 1/(2x)
-> (1), when x-y =0
         x-{1/2x)=0
multiply by 2x,
2x^2 - 1 = 0
       x = (1/ surd 2) or (-1/surd 2)
->    y = (surd 2/2) or (-surd 2 /2)
     
       or  
-> (1), when x+y = 0 (ignored)  ---> x ,y are Real

不肯定！希望能帮到你!

[ 本帖最后由 文锋 于 12-6-2009 07:42 PM 编辑 ]

Ivanlsy

原帖由 文锋 于 11-6-2009 11:58 PM 发表


Given : ( x +iy)^2=i
(x^2 - y^2) + i (2x) = i
so,x^2 - y^2= 0 ------ (1)
     2x = 1 -------------- (2)
       x = 1/2
-> (1), y = 1/2 or -1/2

不肯定！希望能帮到你!

错误展开~ 请再次检查~

meiling

应该是这样吧。。
( x +iy)^2=i
x^2+2xyi+y^2=i
后面自己做，我懒惰

peaceboy

是 x^2+2xyi-y^2=i 吧

junchung2003

原帖由 peaceboy 于 12-6-2009 11:38 AM 发表
是 x^2+2xyi-y^2=i 吧

没讲清楚哦

应该是
-> ( x +iy)^2=i
-> (x+iy)(x+iy)=i
-> x^2 + 2ixy + i^2(y^2) = i  -> i^2 = -1
-> x^2 +2ixy - y^2 = i
-> (x^2 - y^2) + 2ixy - i = 0
-> (x^2 - y^2) + i(2xy - 1) = 0

这边分两part

x^2 - y^2 = 0 和 2xy - 1 = 0

我们先做 x^2 - y^2 = 0
x^2 - y^2 = 0
x^2 = y^2
x = y --(2)

把(2)代入2xy - 1 = 0
2x(x) = 1
x^2 = 1/2
x = √1/2 or -√1/2

Conclusion:
if x = √1/2 then y = √1/2
if x = -√1/2 then y = -√1/2