# Polys / Complex numbers

• Dec 1st 2007, 02:21 PM
Robert G
Polys / Complex numbers
Hello everybody,

I have some problems to do and could use some help. It's all about polynomials and complex numbers, and I basically just started the topic and don't have too much of a clue about it. Any help would be appreciated, but if you could give me some explanations of what to do, how to do it and why to do it, that would be really great.

#1 |(x/(8-x)] <= 2 I need to find the exact values for x. Unfortunately I never had inequalities. Please help me with this.

#2 This one is very hard, please help me with it. x³-x+1=0 has the roots a and b, I need to show that x³+x²-1=0 has the root ab. I also got the hint that I should use x³-x+1=(x-a)(x-b)(x-c) but that doesn't help me too much.

#3 |z+16| = 4|z+1| I need to find |z| if z is a complex number

#4 I need to find the cube roots of -27.

Thank you very very much.

Regards and best wishes,

Robert
• Dec 2nd 2007, 06:07 AM
topsquark
Quote:

Originally Posted by Robert G
#1 |(x/(8-x)] <= 2 I need to find the exact values for x. Unfortunately I never had inequalities. Please help me with this.

Get everything on one side of the inequality:
$\frac{x}{8 - x} - 2 \leq 0$

$\frac{x - 2(8 - x)}{8 - x} \leq 0$

$\frac{3x - 16}{8 - x} \leq 0$

Now find the critical points. These are where either the numerator or denominator are 0. I get $x = \frac{16}{3}, 8$ as critical points.

Now break the number line into sections based on these critical points:
$\left ( -\infty, \frac{16}{3} \right )$
and
$\left ( \frac{16}{3} , 8 \right )$
and
$(8, \infty)$

Now test the inequality on each of these intervals:
$\left ( -\infty, \frac{16}{3} \right )$: $\frac{3x - 16}{8 - x} \leq 0$? Yes.

$\left ( \frac{16}{3} , 8 \right )$: $\frac{3x - 16}{8 - x} \leq 0$? No.

$(8, \infty)$: $\frac{3x - 16}{8 - x} \leq 0$? Yes.

So the final solution set is $\left (-\infty, \frac{16}{3} \right ] \cup (8, \infty)$. (I have included the 16/3 point because this does also satisfy the inequality. If we had < in the inequality we would not do this. We can't include 8 in the solution set because the original expression doesn't exist at x = 8.)

You could also see this by graphing. See the graph below.

-Dan
• Dec 2nd 2007, 06:14 AM
topsquark
Quote:

Originally Posted by Robert G
#4 I need to find the cube roots of -27.

Recall that
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
(If you don't remember this, just multiply it out.)

So
$x^3 = -27$

$x^3 + 27 = 0$

$x^3 + 3^3 = 0$

$(x + 3)(x^2 - 3x + 9) = 0$

Set each factor to 0 and solve.

-Dan
• Dec 2nd 2007, 06:26 AM
topsquark
Quote:

Originally Posted by Robert G
#3 |z+16| = 4|z+1| I need to find |z| if z is a complex number

This won't be the most efficient way, I am sure.

Let $z = a + ib$
where a and b are real.

Then
$|z + 16| = \sqrt{(z + 16)(z^* + 16)}$

$= \sqrt{(a + ib + 16)(a - ib + 16)}$

Similarly
$|z + 1| = \sqrt{(a + ib + 1)(a - ib + 1)}$

Thus we have
$\sqrt{(a + ib + 16)(a - ib + 16)} = 4\sqrt{(a + ib + 1)(a - ib + 1)}$<-- Square both sides

$(a + ib + 16)(a - ib + 16) = 16(a + ib + 1)(a - ib + 1)$

$a^2 + 32a + 256 + b^2 = 16(a^2 + 2a + 1 + b^2)$

$15a^2 + 15b^2 - 240 = 0$

$15(a^2 + b^2) = 240$

$a^2 + b^2 = 16$ <-- $|z| = \sqrt{a^2 + b^2}$

$|z|^2 = 16$

$|z| = 4$

-Dan
• Dec 2nd 2007, 12:25 PM
Robert G
Hey guys, thanks so far, I will look through what you posted in a few minutes, but I was just working on another problem where I got stuck:

(2-3i)/2a+bi = 3+2i I need to find rationals a and b.

But I got stuck at this: 2=6a+2b+4ai+6b+3i

• Dec 2nd 2007, 12:49 PM
Robert G
Quote:

Originally Posted by topsquark
Recall that
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
(If you don't remember this, just multiply it out.)

So
$x^3 = -27$

$x^3 + 27 = 0$

$x^3 + 3^3 = 0$

$(x + 3)(x^2 + 3x + 9) = 0$

Set each factor to 0 and solve.

-Dan

Thank you, but how exactly do I solve this? I get -3 as the only root, but there needs to be an irrational root, how do I get this? Thanks.
• Dec 2nd 2007, 01:28 PM
topsquark
Quote:

Originally Posted by Robert G
Thank you, but how exactly do I solve this? I get -3 as the only root, but there needs to be an irrational root, how do I get this? Thanks.

Hint: You only set one of your factors equal to 0.

Now solve
$x^2 - 3x + 9 = 0$ <-- The solution is two complex numbers.

(There was a typo in my original post. The sign on the 3x term was + instead of -.)

-Dan
• Dec 2nd 2007, 01:32 PM
topsquark
Quote:

Originally Posted by Robert G
Hey guys, thanks so far, I will look through what you posted in a few minutes, but I was just working on another problem where I got stuck:

(2-3i)/2a+bi = 3+2i I need to find rationals a and b.

But I got stuck at this: 2=6a+2b+4ai+6b+3i

This is
$\frac{2 - 3i}{2a + bi} = 3 + 2i$?

$2 - 3i = (2a + bi)(3 + 2i)$

$2 - 3i = 6a + 4ai + 3bi - 2b$

Now equate reals to reals and imaginaries to imaginaries:
$2 = 6a - 2b$

$-3 = 4a + 3b$

Now solve the system of equations.

-Dan
• Dec 2nd 2007, 01:39 PM
Robert G
Quote:

Originally Posted by topsquark
Recall that
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
(If you don't remember this, just multiply it out.)

So
$x^3 = -27$

$x^3 + 27 = 0$

$x^3 + 3^3 = 0$

$(x + 3)(x^2 - 3x + 9) = 0$

Set each factor to 0 and solve.

-Dan

Quote:

Originally Posted by topsquark
Hint: You only set one of your factors equal to 0.

Now solve
$x^2 - 3x + 9 = 0$ <-- The solution is two complex numbers.

(There was a typo in my original post. The sign on the 3x term was + instead of -.)

-Dan

Thanks, but shouldn't it then be x²-6x+9 ? Because of (a+b)² = a²+2ab+b² ? I mean 2*3*x=6x ?

And how do I get complex solutions, I don't get a complex solution. I get (if I take 6x instead of 3) 25.5 and -19.5. How would it be correct?

And by the way, has someone an idea for the second problem? I think it's the hardest and I have no clue.

But in any case, thank you a lot.
• Dec 2nd 2007, 06:01 PM
topsquark
Quote:

Originally Posted by Robert G
Thanks, but shouldn't it then be x²-6x+9 ? Because of (a+b)² = a²+2ab+b² ? I mean 2*3*x=6x ?

I wrote it correctly.
$a^3 + b^3 \neq (a + b)^3$

So we have
$x^3 + 27 = 0$

$(x + 3)(x^2 - 3x + 9) = 0$

$x + 3 = 0 \implies x = -3$
or
$x^2 + 3x + 9 = 0$

$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 9}}{2}$

$x = -\frac{3}{2} \pm i \frac{3 \sqrt{3}}{2}$

-Dan