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Conjugates (1 Viewer)

Calculon

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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

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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
 

Calculon

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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

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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

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That's just a property. :p
(a + ib) (a - ib) = a<sup>2</sup> + b<sup>2</sup>

An easy thing to use might be the fact that solutions to (real) polynomials like x<sup>2</sup> + x + 5 = 0 are (complex) conjugate pairs. They'll multiply to give a real and these conjugates are always a + ib and a - ib.
 

KeypadSDM

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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) = a<sup>2</sup> - c<sup>2</sup>b

(a + cSqrt) + (a - cSqrt) = 2a
 

Xayma

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

KeypadSDM

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As a more general equation:

To find a polynomial with a root:

a<sup>1/2</sup> - b<sup>1/2</sup>

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

f(x) = x<sup>4</sup> - 2(a + b)x<sup>2</sup> + (a - b)<sup>2</sup>
 

Euler

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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

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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'? :p

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

Euler

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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.
 

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