# Complex Numbers - Conjugates

1. Sep 30, 2004

### JasonRox

It says that:

x^2=the conjugate squared, which is:

(a+bi)^2=(a-bi)^2

How can I show this?

This isn't homework or anything, but I don't get where they got this from.

I'm reluctant to move on until this is solved or understood.

2. Sep 30, 2004

### JasonRox

Now I'm at:

a^2+2bi-b^2=a^2+2b(-i)-b^2

3. Sep 30, 2004

### JasonRox

If I square both sides, I now get:

a^4-4b^2-b^4=a^4-4b^2-b^4

They equal.

Is this correct?

4. Sep 30, 2004

### vsage

You're trying to prove (a+bi)^2 = (a+b*(-i))^2? If so then they're not. If i'm not mistaken though |a+bi| = |a-bi| When you squared both sides you did something a little off. I didn't look too closely but solving out (2+3i)^2 and (2-3i)^2 you find they are different numbers of the same magnitude.

Last edited by a moderator: Sep 30, 2004
5. Oct 1, 2004

### Tide

Isn't it obvious that b = 0?

6. Oct 1, 2004

### matt grime

perhaps you're trying to show that xx*=|x|^2
where x* is the conjugate.

what yo'ure trying to prove isn't true, and your algebra is wrong (in squaring both sides you've managed to lose the i some how)

7. Oct 1, 2004

### Zurtex

I may be getting horribly confused here but from the original post it seems to you are to prove:

$$z^2 = (z^*)^2$$

Where z is any complex number?

$$(a + bi)^2 = a^2 + 2abi + b^2$$

$$(a - bi)^2 = a^2 - 2abi + b^2$$

Quite clearly it is not always true that:

$$a^2 + 2abi + b^2 = a^2 - 2abi + b^2$$

In fact it is only true when a and/or b are 0.

8. Oct 1, 2004

### BobG

I'm not sure what you're trying to show, either. The conjugate of (a+bi) is (a-bi), but they are not equal to each other. The purpose of the conjugate is to remove your imaginary portion from the denominator (the same method can be used to remove your square roots from the denominator).

If you multiply (a+bi) and (a-bi), you get: a^2 +abi - abi + b^2. The +abi and -abi cancel out, leaving you with a^2-b^2. Your denominator is left with all real numbers.

You can do the same with a denominator such as (3+sqrt(3)). Multiply both the numerator and denominator by (3-sqrt(3)) and your denominator reduces to (9-3), or 6.

9. Oct 1, 2004

### Gonzolo

Does it say x^2 ?
or does it say |x|^2 ?

10. Oct 1, 2004

### HallsofIvy

Does WHAT say x2 or |x|2???

It is not at all clear whether you are trying to solve the equation x2= (conjugate of x)2 (which is satisfied by the set of all real numbers union the set of all imaginary numbers: a+0i or 0+ bi) or whether you were asserting that
"x2= x times conjugate of x) (if that is what you are saying, then no, you need "|x|2= x times conjugate of x". x2 is just x times x, in the complex numbers as well as the real numbers.).

11. Oct 1, 2004

### JasonRox

Let me change it up.

The conjugate is found by simply changing i to -i. Right?

So...

$$(a+bi)^2=(a+b(-i))^2$$

If you solve this,you get...

$$a^2+2abi-b^2=a^2+2ab(-i)-b^2$$

Also note...

$$i^2=(-i)^2=-1$$

NOTE: Where was my algebra wrong? You guys came up with +b^2, and that's wrong because it is actually...

$$b^2*(-i)^2$$

From above you see that (-i)^2=-1. Thus...

$$b^2*(-i)^2 = -b^2$$

I just showed that it is true.

Matt Grime, you lose the i because i^2=-1.

Note: I could be wrong. I will check over the book again. It is not lx*l^2.

12. Oct 3, 2004

### matt grime

You are wrong. You lost *all* the i's in what you wrote

"a^4-4b^2-b^4=a^4-4b^2-b^4"

going from post 2 to post 3.
It really won't do for you to play games with me: your logic in going from post 2 to post 3 presumes you to be in char 2. you've done the age old idiotic trick of writing:

(x+y)^2=x^2+y^2

you wanna go check?

13. Oct 3, 2004

### JasonRox

I converted the i^2 into -1.

How else did I get the negatives??? The i's didn't get lost. Watch it carefully.

I did not do the age old trick.

Look at it!

14. Oct 3, 2004

### JasonRox

I'll go slower for you guys.

$$(a+bi)^2=(a+b(-i))^2$$
$$a^2+2abi+b^2i^2=a^2+2ab(-i)+b^2(-i)^2$$

If you know the basics of Complex Numbers, we know that...

$$i^2=(-i)^2=-1$$

Taking that rule and applying it to...

$$a^2+2abi+b^2i^2=a^2+2ab(-i)+b^2(-i)^2$$

we get...

$$a^2+2abi+b^2(-1)=a^2+2ab(-i)+b^2(-1)$$

to...

$$a^2+2abi-b^2=a^2+2ab(-i)-b^2$$

Now, I square both sides and we get...

$$a^4+4a^2b^2i^2-b^2=a^4+4a^2b^2(-i)^2-b^4$$

Taking the SAME rule...

$$i^2=(-i)^2=-1$$

we get...

$$a^4-4a^2b^2-b^2=a^4-4a^2b^2-b^4$$

Now, they equal!

I can't believe you didn't know where the i's went. This is the stuff they teach you in the first chapters.

Note: I could have made a mistake, which is normal. If I did, please POST the mistake and not some non-sense that doesn't help because I am doubting whether you guys seen i before.

15. Oct 3, 2004

### arildno

The last "equation" is not an equation.
It says:
2abi=-2abi
which is untrue.
Hence, your first "equation" isn't any equation either.

16. Oct 3, 2004

### gerben

This step is not right

$$(a + b + c)^2=a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$$

.

17. Oct 3, 2004

### JasonRox

Oh ok.

It still equals.

18. Oct 3, 2004

### gerben

No it does not.

on the left you have:
$$(a + b - c)^2=a^2 + b^2 + c^2 + 2ab - 2ac - 2bc$$
and on the right:
$$(a - b - c)^2=a^2 + b^2 + c^2 - 2ab - 2ac + 2bc$$

$$a^2 + b^2 + c^2 + 2ab - 2ac - 2bc \neq a^2 + b^2 + c^2 - 2ab - 2ac + 2bc$$

the i is only in the b term so you can change the sign in the terms in which b is squared, doing so gives:

$$a^2 - B^2 + c^2 + 2ab - 2ac - 2bc \neq a^2 - B^2 + c^2 - 2ab - 2ac + 2bc$$
(using: $$b = Bi^2$$)

19. Oct 3, 2004

### JasonRox

Also, it says that lxl^2=a^2+b^2.

Can some explain this a little?

20. Oct 4, 2004

### TenaliRaman

|x|^2 = xx*
where x* is complex conjugate of x
so if x = a+ib x* = a-ib
|x|^2 = xx* = (a+ib)(a-ib) = a^2+b^2

-- AI