What is the relationship between complex numbers and conjugates?

[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.

 

[JasonRox]

Now I'm at:

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

 

[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?

 

[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.

 

[Tide]

Isn't it obvious that b = 0?

 

[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)

 

[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.

 

[BobG]

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.

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.

 

[Gonzolo]

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

 

[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.).

 

[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.

 

[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 want to go check?

 

[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!

 

[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.

 

[arildno]

I'll go slower for you guys.

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

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

.

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

 

[gerben]

This step is not right

...
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
...

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

.

 

[JasonRox]

Oh ok.

It still equals.

 

[gerben]

Oh ok.

It still equals.

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)

 

[JasonRox]

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

Can some explain this a little?

 

[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

 

[matt grime]

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.

post two contains an equation.

post three you allegedly "square both sides" of the equation


actually you don't do any such thing.

and i don't think you want to start comparing qualifications.

 

[JasonRox]

and i don't think you want to start comparing qualifications.

Of course not. :)

I know that lxl^2=xx*, but how do I show this by breaking up both sides?

This is what I am asking, and I need your help.

 

[matt grime]

That is now what you're asking, that isn't what you were asking before, and it's easy. In fact wasn't this the thing I asked if you meant to ask in my first reply and that you ignored?

What is |x| where x=a+ib?
hence what is |x|^2?
Now, what is (a+ib)(a-ib) which is xx*?

Of course if you don't know that |x| = sqrt(a^2+b^2) then you're in trouble, but that's the definition, so you ought to know it.

 

[JasonRox]

That is the non-negative real numbers or the modulus, right?

Is this just to express lxl as sqrt(a^2+b^2)?

If that's all it is, I'm good. For now, anyways.

Note: I really did mess up this post. :s

 

[matt grime]

"That is the non-negative real numbers or the modulus, right?"

What is the non-negative real numbers or the modulus?

The definition of the modulus of a complex number a+ib with a and b real is sqrt(a^2+b^2). It is the length of the line segment in the complex plane from the origin to a+ib, it is just pythagoras.

 

[JasonRox]

Exactly.

They mean the same thing.

I also know that it is just the phythagorean theorem.

Also, I did not get into the geometry yet because that's chapter 2.

Thanks, for going through all the trouble.