# Complex Number finding the roots

• Nov 3rd 2010, 01:48 AM
Paymemoney
Complex Number finding the roots
Hi

I need help on the following equation:

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

$u=z^2$
$u^2+2u+4=0$

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
• Nov 3rd 2010, 02:51 AM
Prove It
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 $\displaystyle z$.
• Nov 3rd 2010, 06:55 AM
HallsofIvy
Quote:

Originally Posted by Paymemoney
Hi

I need help on the following equation:

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

$u=z^2$
$u^2+2u+4=0$

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!
• Nov 3rd 2010, 08:46 AM
Soroban
Hello, Paymemoney!

Your arithmetic is off . . .

Quote:

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

• Nov 3rd 2010, 01:33 PM
Paymemoney
• Nov 3rd 2010, 02:29 PM
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

$u=z$

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

what would you do next?
• Nov 3rd 2010, 02:34 PM
mr fantastic
Quote:

Originally Posted by Paymemoney
Hi

I need help on the following equation:

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

$u=z^2$
$u^2+2u+4=0$

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.
• Nov 3rd 2010, 02:40 PM
mr fantastic
Quote:

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.