Conjugates (1 Viewer)

[Calculon]

The definition of a conjugate is that it is something you multiply the original number by to make it rational/real right? Hence the conjugate of 1+i is 1-i, but if you think about it, it can also be -1+i
If this is true, how come textbooks tell you that
z + conj.(z) = 2 Re(z)

Because according to my calculations that's just not true

EDIT: Yes I am going to play along with the way the book says, i don't want to lose marks over it.

 

[Heinz]

Originally posted by Calculon
The definition of a conjugate is that it is something you multiply the original number by to make it rational/real right? Hence the conjugate of 1+i is 1-i, but if you think about it, it can also be -1+i
If this is true, how come textbooks tell you that
z + conj.(z) = 2 Re(z)

Because according to my calculations that's just not true

EDIT: Yes I am going to play along with the way the book says, i don't want to lose marks over it.

WAH? the conjugates just 1-i so z + conj. z = 2 Re z i.e. 1 + i + 1 - 1 = 2

 

[:: ck ::]

ermmm u dont make ne sense

 

[BlackJack]

The definition of a conjugate is:
if z= x + iy
then, conj. (z) = x - iy

You add conjugates to obtain a real number.

 

[Calculon]

I wasnt taught that way i was just told that a conjugate is a number which when you multiply the original number by it it will be rational/real, and i was given a couple of examples

 

[Xayma]

Well if you multiply them you also get a real number, conjugate surds will give a rational number when multiplied but not added so maybe they taught it that way so as not to confuse you to much.

 

[BlackJack]

That's just a property.
(a + ib) (a - ib) = a² + b²

An easy thing to use might be the fact that solutions to (real) polynomials like x² + x + 5 = 0 are (complex) conjugate pairs. They'll multiply to give a real and these conjugates are always a + ib and a - ib.

 

[kimmeh]

all real roots occur in conjugate pairs?

 

[freaking_out]

Originally posted by kimmeh
all real roots occur in conjugate pairs?

yep- but thats if the equation coefficients are rational integers (or something like that).

 

[kimmeh]

yeah all real roots occur in conjugate pairs

 

[KeypadSDM]

Originally posted by kimmeh
yeah all real roots occur in conjugate pairs

Well they do.

It's just that the pairs are the same number.

 

[KeypadSDM]

Originally posted by Xayma
conjugate surds will give a rational number when multiplied but not added so maybe they taught it that way so as not to confuse you to much.

Uhhh

(a + cSqrt)(a - cSqrt) = a² - c²b

(a + cSqrt) + (a - cSqrt) = 2a

 

[Xayma]

What if both are surds? If they are multiplied it will give rational if added you will still get a surd.

 

[KeypadSDM]

Wait a second. Does this mean that Sqrt[3] - Sqrt[2] is trancedental ...

Ponders thought ...

 

[KeypadSDM]

Nah, I was wrong

That's a solution of

f(x) = x⁴ - 10x² + 1

 

[KeypadSDM]

As a more general equation:

To find a polynomial with a root:

a^1/2 - b^1/2

Where a and b are non square integers, the equation is:

f(x) = x⁴ - 2(a + b)x² + (a - b)²

 

[Euler]

There is a real mathematical definition of what a conjugate is.

For what you need, though, it is enough to remember that the conjugate of a+ib is always a-ib.

For those that are interested: for a complex number a+ib, there is an irreducible real polynomial which has a-ib as a zero. The other roots of the polynomial are the conjugates.

If you are working with surds, then you will be looking at irreducible rational polynomials.

 

[Affinity]

Originally posted by Euler

For those that are interested: for a complex number a+ib, there is an irreducible real polynomial which has a-ib as a zero. The other roots of the polynomial are the conjugates.

you mean 'which has a+ib as a zero'?

I think it's better to use a-ib as definition and take this as a propety though

 

[Euler]

my bad...thanks Affinity.

I do mean a+ib as a zero.

That is the definition of conjugate, but you can just remember that for complex numbers, the conjugate of a+ib is always a-ib.