Complex Number finding the roots

**Paymemoney** · November 3rd 2010, 01:48 AM

Hi

I need help on the following equation:

1) $z^4+2z^2+4 =0$

$u=z^2$
$u^2+2u+4=0$
use quadratic equation:

i get

$u= \frac{-2 +\sqrt{16}}{4}$

or

$u= \frac{-2 -\sqrt{16}}{4}$

my question is do i simplify this to find the roots or leave it?

P.S

**Prove It** · November 3rd 2010, 02:51 AM

I think completing the square might be better in this case...

$\displaystyle z^4 + 2z^2 + 4 = 0$

$\displaystyle z^4 + 2z^2 + 1^2 - 1^2 + 4 = 0$

$\displaystyle (z^2 + 1)^2 + 3 = 0$

$\displaystyle (z^2 + 1)^2 = -3$

$\displaystyle z^2 + 1 = \pm \sqrt{3} i$

$\displaystyle z^2 = -1 \pm \sqrt{3} i$

From here, you should convert to polars to find the four values of $\displaystyle z$ .

**HallsofIvy** · November 3rd 2010, 06:55 AM

Originally Posted by Paymemoney

Hi

I need help on the following equation:

1) $z^4+2z^2+4 =0$

$u=z^2$
$u^2+2u+4=0$
use quadratic equation:

i get

$u= \frac{-2 +\sqrt{16}}{4}$

or

$u= \frac{-2 -\sqrt{16}}{4}$

my question is do i simplify this to find the roots or leave it?

P.S

You are aware that $\sqrt{16}= 4$ aren't you? I can't imagine not using that simplification!

**Soroban** · November 3rd 2010, 08:46 AM

Hello, Paymemoney!

Your arithmetic is off . . .

1) Solve: . $z^4+2z^2+4 \:=\:0$

Let: $u=z^2$

Substitute: $u^2+2u+4\:=\:0$

Use Quadratic Formula: . $u= \dfrac{-2 +\sqrt{16}}{4}$ .no

Did you forget the Quadratic Formula?
For $ax^2 + bx + c \:=\:0$ the roots are: . $x \;=\;\dfrac{-b \pm\sqrt{b^2-4c}}{2a}$

$\text{We have: }\:u^2 + 2u + 4 \:=\:0 \quad\Rightarrow\quad \begin{Bmatrix}a\:=\:1\\b\:=\:2\\c\:=\:4 \end{Bmatrix}$

$\text{Hence: }\;u \;=\;\dfrac{\text{-}2 \pm\sqrt{2^2-4(1)(4)}}{2(1)} \;=\;\dfrac{\text{-}2\pm\sqrt{\text{-}12}}{2}$

. . . . . . $u \;=\;\dfrac{\text{-}2\pm2\sqrt{\text{-}3}}{2} \;=\;\text{-}1 \pm i\sqrt{3}$

Back-substitute: . $z^2 \;=\;\text{-}1\pm i\sqrt{3}$

. . Therefore: . $z \;=\;\pm\sqrt{\text{-}1 \pm i\sqrt{3}}$

**Paymemoney** · November 3rd 2010, 01:33 PM

oops yeh my bad, my quadratic equation was the mistake

**Paymemoney** · November 3rd 2010, 02:29 PM

Just a quick question if i had something like this how would i find the roots

$z^4+2z^3+5z^2+8z+4=0$
This is what i have done

$u=z$

$u^3+12u^2+5u+8u+4=0$
$u^3+12u^2+13u+4=0$

what would you do next?

**mr fantastic** · November 3rd 2010, 02:34 PM

Originally Posted by Paymemoney

Hi

I need help on the following equation:

1) $z^4+2z^2+4 =0$

$u=z^2$
$u^2+2u+4=0$
use quadratic equation:

i get

$u= \frac{-2 +\sqrt{16}}{4}$

or

$u= \frac{-2 -\sqrt{16}}{4}$

my question is do i simplify this to find the roots or leave it?

P.S

Your solutions for u are wrong.

**mr fantastic** · November 3rd 2010, 02:40 PM

Originally Posted by Paymemoney

Just a quick question if i had something like this how would i find the roots

$z^4+2z^3+5z^2+8z+4=0$
This is what i have done

$u=z$ Mr F says: What does this achieve?

$u^3+12u^2+5u+8u+4=0$ Mr F says: This is totally wrong.
$u^3+12u^2+13u+4=0$

what would you do next?

It's very clear that z = -1 is a solution to $z^4+2z^3+5z^2+8z+4=0$ . Therefore z + 1 is a factor. Divide z + 1 into $z^4+2z^3+5z^2+8z+4$ to get a cubic factor. It is very clear that z + 1 is also a factor of the cubic. Divide it into the cubic to get a quadratic factor. Factorise the quadratic in the usual way.

From the factor you get the solutions. There is sure to be examples like this in your class notes and textbook. Go back and review this material.

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