数学Paper 1讨论专区

Lov瑜瑜4ever

做法都是差不多一样的
我是先去除A，D，C的unknown
然后再来找剩下的
找到了E才找到B的。。

破晓时分

a) if z=a+bi and l z+i-1 l = 2, show that a^2 + b^2 -2a+2b-2=0

b) Obtain the quadratic equation with the roots -3+4i and -3-4i in the general form.

c) Given z^2=-3+4i. Find the complex numbers of z in the form x+yi.

solution..please..thanks..

Lov瑜瑜4ever

a) if z=a+bi and l z+i-1 l = 2, show that a^2 + b^2 -2a+2b-2=0

b) Obtain the quadratic equation w ...
破晓时分 发表于 23-6-2010 08:27 PM

a) if z=a+bi and l z+i-1 l = 2, show that a^2 + b^2 -2a+2b-2=0


l z+i-1 l = 2
|a+bi+i-1|=2
|(a-1)+(b+1)i|=2
[(a-1)^2+(b+1)^2]^1/2=2
a^2-2a+1+b^2+2ab+1=4
a^2+b^2-2a+2b-2=0 (Shown)

Lov瑜瑜4ever

a) if z=a+bi and l z+i-1 l = 2, show that a^2 + b^2 -2a+2b-2=0

b) Obtain the quadratic equation w ...
破晓时分 发表于 23-6-2010 08:27 PM

b) Obtain the quadratic equation with the roots -3+4i and -3-4i in the general form.

Let A and B is the root of that equation
so, A+B=(-3+4i)+(-3-4i)=-6
     AB=(-3+4i)(-3-4i)=(3-4i)(3+4i)=3^2-(4i)^2=9+16=25

The general form,  x^2-(A+B)x+AB=0  (This thing you have learn since form4)
                          x^2+6x+25=0

Lov瑜瑜4ever

a) if z=a+bi and l z+i-1 l = 2, show that a^2 + b^2 -2a+2b-2=0

b) Obtain the quadratic equation w ...
破晓时分 发表于 23-6-2010 08:27 PM

c) Given z^2=-3+4i. Find the complex numbers of z in the form x+yi.

Let z=x+yi
so, (x+yi)^2=-3+4i
      x^2-y^2+2xyi=-3+4i

By comparison,
x^2-y^2=-3-----(1)
2xyi=4i
  xy=2------(2)
   y=2/x-------(3)

substitute (3) into (1),
............
接下来你自己做了
下面的很麻烦><
paiseh...

破晓时分

a) if z=a+bi and l z+i-1 l = 2, show that a^2 + b^2 -2a+2b-2=0


l z+i-1 l = 2
|a+bi+i-1|=2
...
Lov瑜瑜4ever 发表于 23-6-2010 08:56 PM

|(a-1)+(b+1)i|=2
[(a-1)^2+(b+1)^2]^1/2=2


为什么会出现square呢?然后那个1/2为什么会跳出来？

Lov瑜瑜4ever

|(a-1)+(b+1)i|=2
[(a-1)^2+(b+1)^2]^1/2=2


为什么会出现square呢?然后那个1/2为什么会跳出来？
破晓时分 发表于 23-6-2010 09:04 PM

modulus of z can be written as |z|,rite?
if z=(a-1)+(b+1)i,
then |z|=|(a-1)+(b+1)i|
and question said |(a-1)+(b+1)i|=2
the solving method is same as how u find modulus...

破晓时分

回复 1267# Lov瑜瑜4ever


   okay..thank you

Lov瑜瑜4ever

回复  Lov瑜瑜4ever


   okay..thank you
破晓时分 发表于 23-6-2010 09:15 PM

welcome...

糖果candy

幫我。。
find the four zeros of p(x)

p(x)=(x+1) (5x-3) (5x^2 +2x+1)

Lov瑜瑜4ever

幫我。。
find the four zeros of p(x)

p(x)=(x+1) (5x-3) (5x^2 +2x+1)
糖果candy 发表于 24-6-2010 10:38 PM

4 zeros 指的是4个x的答案吗？
我有点忘记了><

糖果candy

沒錯              =)

Lov瑜瑜4ever

幫我。。
find the four zeros of p(x)

p(x)=(x+1) (5x-3) (5x^2 +2x+1)
糖果candy 发表于 24-6-2010 10:38 PM

我想起来了
很容易做的
就是(x+1)(5x-3)(5x^2+2x+1)=0
所以 x+1=0,x=-1 ; 5x-3=0,x=3/5
5x^2+2x+1=0
x=(-1+2i)/5 or x=(-1-2i)/5
4个x的答案就找到了
(-16)^1/2=[16(-1)]^1/2=[(4^2)(-1)]^1/2=4i

糖果candy

我不知道怎樣從
5x^2+2x+1=0  變去
x=(-1+2i)/5 or x=(-1-2i)/5

Lov瑜瑜4ever

我不知道怎樣從
5x^2+2x+1=0  變去
x=(-1+2i)/5 or x=(-1-2i)/5
糖果candy 发表于 24-6-2010 10:54 PM

from the eqn, a=5,b=2,c=1
so b^2-4ac=2^2-4(5)(1)=-16
x=[-b+(b^2-4ac)^1/2]/2a  or  x=[-b-(b^2-4ac)^1/2]/2a
substitute the value of a,b,c and b^2-4ac, eventually u will get the answer...

Allmaths

回复 1270# 糖果candy[img][/img]

Allmaths

第一次用MathType..打到好辛苦.....walrein_lim88 还真强...打到很快...

Lov瑜瑜4ever

Picture is the most clear expression...

Lov瑜瑜4ever

第一次用MathType..打到好辛苦.....walrein_lim88 还真强...打到很快...
Allmaths 发表于 24-6-2010 11:02 PM

what is MathType?
Is this a software?

Allmaths

回复 1279# Lov瑜瑜4ever


是的..不过只给你30天trial...