Please correct my working for: Factorise z^3 +1 over C.

**anees7112** · February 5th 2011, 09:09 PM

Hello

The question is Factorise z^3 +i over C . (Book's answer is (z-i)(z^2+iz-1))

I have attempted it 3 ways: Number 1 seems to work, but cannot get 2 or 3 to work.

Sorry I didnt know how to do long division in word so did it by hand instead.

In (1) I tried to divide z^3+i by (z-i) which I think is a factor
and this happened to give the same answer as the book.

In (2) I tried to divide z^3+i by (z+i) which I think is also a factor but I couldnt not complete the division properly.

In (3) I attempted to use this method :

but I do know now how to this method with this question.

Could you please correct my working and show me how to use method 3 if possible as I think that is the method the teacher wants us to do for these types of questions.

Thanks

*note: should have read + i not +1 in title.

**TheCoffeeMachine** · February 5th 2011, 09:14 PM

Originally Posted by anees7112

In (2) I tried to divide z^3+i by (z+i) which I think is also a factor but I couldnt not complete the division properly.

Why do think that is a factor? Hint: it isn't. The third method fails for the same reason.

**TheEmptySet** · February 5th 2011, 09:17 PM

first notice that

$z^3+i=z^3-i^3$ Remember that $i^2=-1$

Now this is the difference of cubes

$(a^3-b^3)=(a-b)(a^2+ab+b^2)$

$z^3-i^3=(z-i)(z^2+iz+i^2)=z^3-i^3=(z-i)(z^2+iz-1)$

**anees7112** · February 5th 2011, 09:30 PM

Originally Posted by TheCoffeeMachine

Why do think that is a factor? Hint: it isn't. The third method fails for the same reason.

Hello Thecoffeemachine.

I thought that if z-i was a factor z+i would be too instead of if z-i is a root z+i is a root too, silly me.

Thank you for clearing that up

and

Hi TheEmptySet

Originally Posted by TheEmptySet

first notice that

$z^3+i=z^3-i^3$ Remember that $i^2=-1$

Now this is the difference of cubes

$(a^3-b^3)=(a-b)(a^2+ab+b^2)$

$z^3-i^3=(z-i)(z^2+iz+i^2)=z^3-i^3=(z-i)(z^2+iz-1)$

Thank you for showing me how to do it that way

**TheCoffeeMachine** · February 5th 2011, 09:45 PM

Originally Posted by anees7112

I thought that if z-i was a factor z+i would be too instead of if z-i is a root z+i is a root too, silly me.

I think you're still missing something here. $z = i$ is a root of course, and consequently $z-i$ is a factor. But $z = -i$ isn't root, and $z+i$ isn't a factor either. I think the mistake comes from the fact that, for any polynomial $p(x)$ with real coefficients, if $x = z$ is complex solution that satifies $p(x)$ , then so is $x = -z$ ; thus both $(x-z)$ and $(x+z)$ are factors of $p(x)$ . But this isn't the case here because not all the coefficients of our polynomial are real; i.e. our constant term is $i$ . Thus the rule does not apply in this case. (I think this is pretty common mistake, so it isn't silly at all).

**anees7112** · February 5th 2011, 10:03 PM

Originally Posted by TheCoffeeMachine

I think you're still missing something here. $z = i$ is a root of course, and consequently $z-i$ is a factor. But $z = -i$ isn't root, and $z+i$ isn't a factor either. I think the mistake comes from the fact that, for any polynomial $p(x)$ with real coefficients, if $x = z$ is complex solution that satifies $p(x)$ , then so is $x = -z$ ; thus both $(x-z)$ and $(x+z)$ are factors of $p(x)$ . But this isn't the case here because not all the coefficients of our polynomial are real; i.e. our constant term is $i$ . Thus the rule does not apply in this case. (I think this is pretty common mistake, so it isn't silly at all).

ah okay. Thanks for clearing that up for me, I hadn't understood that before

**Archie Meade** · February 6th 2011, 03:22 AM

Originally Posted by anees7112

Hello

The question is Factorise z^3 +i over C . (Book's answer is (z-i)(z^2+iz-1))

I have attempted it 3 ways: Number 1 seems to work, but cannot get 2 or 3 to work.

Sorry I didnt know how to do long division in word so did it by hand instead.

In (1) I tried to divide z^3+i by (z-i) which I think is a factor
and this happened to give the same answer as the book.

You are correct, but you should have written above that P(i)=0

In (2) I tried to divide z^3+i by (z+i) which I think is also a factor but I couldnt not complete the division properly.

Your division is correct and if it was a factor, your remainder would be 0.
The remainder is not zero so that is not a factor of the complex polynomial.

In (3) I attempted to use this method :

but I do know now how to this method with this question.

Notice this method states that the coefficients must be real (see "Think 1").

Could you please correct my working and show me how to use method 3 if possible as I think that is the method the teacher wants us to do for these types of questions.

Thanks

*note: should have read + i not +1 in title.

Alternatively

$(z-i)\left(z^2-1)=z^3-iz^2-z+i$

so we need a z term to cancel the extra terms

$(z-i)\left(z^2+kz-1\right)=z^3+kz^2-z-iz^2-ikz+i=z^3+(k-i)z^2-(ik+1)z+i$

$=z^3+(0)z^2+(0)z+i$

for which the coefficient of $z^2$ and $z$ both give

$k=i$

**CaptainBlack** · February 7th 2011, 12:11 AM

Originally Posted by anees7112

Hello Thecoffeemachine.

I thought that if z-i was a factor z+i would be too instead of if z-i is a root z+i is a root too, silly me.

Thank you for clearing that up

If $P(z)$ is a polynomial with real coefficients then if $$$z$ is a root then so is $$$\overline{z}$ , but this fails for your case because:

You do not have a polynomial with real coefficients

CB

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