# Solving an equation with de Moivre's formula

• Jan 30th 2009, 04:31 AM
Greenb
Solving an equation with de Moivre's formula
$z^2=-119+120i$

$z=r(cosv+isinv)$

and

$
-119+120i=(-119+120i)(cos0+isin0)
$

De Moivres states
$
z^2=r^2(cos2v+isin2v)
$

$z^2=-119+120i$

can be written

$
(-119+120i)(cos0+isin0)=r^2(cos2v+isin2v)
$

The two lines equals eachother if

$z^2=-119+120i$
and
$v=0+n*pi$
making the angles
$v1=0$
$v2=pi$

This is my problem:

$z^2=-119+120i$
• Jan 30th 2009, 05:34 AM
Soroban
Hello, Greenb!

Is this the problem?

Quote:

$z^2\:=\:-119+120i$. .Find $z.$
On the argand diagram, we have:
Code:

        *         : * 169     120 :  *         :    * θ     - - + - - - * - - - -           -119
Hence: . $\begin{array}{ccc}\cos\theta &=&\text{-}\frac{119}{169} \\ \\[-4mm] \sin\theta &=& \frac{120}{169} \end{array}$

We have: . $z^2 \;=\;169(\cos\theta + i\sin\theta)$

And we want: . $z \;=\;13\left(\cos\tfrac{\theta}{2} + i\sin\tfrac{\theta}{2}\right)$ .[1]

Using half-angle identities:

. . $\cos\tfrac{\theta}{2} \:=\:\sqrt{\frac{1+\cos\theta}{2}} \:=\:\sqrt{\frac{1-\frac{119}{169}}{2}} \;=\;\sqrt{\frac{50}{338}} \:=\:\frac{5}{13}$

. . $\sin\tfrac{\theta}{2} \:=\:\sqrt{\frac{1-\cos\theta}{2}} \:=\:\sqrt{\frac{1-(\text{-}\frac{119}{169})}{2}} \;=\;\sqrt{\frac{288}{338}} \;=\;\frac{12}{13}$

Substitute into [1]: . $z \;=\;13\left(\frac{5}{13} + i\frac{12}{13}\right) \quad\Rightarrow\quad\boxed{ z\;=\;5 + 12i}$

• Jan 30th 2009, 07:43 AM
Greenb
Ah I forgot that |z| equals the hypotenuse, but I'm not with you all the way on the half angle identities part, seeing that its a cubicequation, it should have two roots.

And what if the equation was;

$z^4=4-4i$

could it be solved the same way?

I'm Swedish by the way so sorry if I don't get the termonoligy right.