# Complex Numbers :D

• Jan 19th 2008, 06:37 AM
brd_7
Complex Numbers :D
The question says,

Find all splutions of the following equations, expressing your answers in Cartesian form (i.e as z=x+iy)

(a) $z^4 - 1 = i\sqrt3$

Do i simple open it out and simplify?.. is there anything else that needs to be done?
• Jan 19th 2008, 06:51 AM
Plato
Find the four fourth roots of $1 + i\sqrt 3$.
• Jan 19th 2008, 07:15 AM
brd_7
Oh ok, And how do i go about starting that?
• Jan 19th 2008, 08:09 AM
topsquark
Quote:

Originally Posted by brd_7
Oh ok, And how do i go about starting that?

I favor putting the complex number into polar form:
$1 + i \sqrt{3} = re^{i \theta}$

Then go from there.

-Dan
• Jan 19th 2008, 12:24 PM
brd_7
$z^4 = 1 + i \sqrt{3}$

Represent $1 + i \sqrt{3}$ in the form $r(\cos \theta + i \sin \theta)$

I got $z^4 = 2 ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} )$

Let $z = r (\cos \theta + i \sin \theta)$

Then $z^4 = r^4 ( cos \theta + i \sin \theta)^4$

So $r^4 ( cos \theta + i \sin \theta)^4 = 2 ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} )$

And then-
z^4 = 2.{cos[(1/3 + 2n)pi] + isin[(1/3 + 2n)pi]}. And so
z = 2^[1/4]{cos[(1/12 + n/2)pi] + isin[(1/12 + n/2)pi]].

Is this all right?.. What do i do now?.. How do i change it back to x +iy?
• Jan 19th 2008, 12:40 PM
Plato
Yes it is correct if this is what you mean.
$z = \sqrt[4]{2}\left[ {\cos \left( {\frac{\pi }{{12}} + \frac{{k\pi }}{2}} \right) + i\sin \left( {\frac{\pi }{{12}} + \frac{{k\pi }}{2}} \right)} \right]\,,\,k = 0,1,2,3
$
• Jan 19th 2008, 02:04 PM
brd_7
and so how do i show it in terms of x+iy?

Thanks for all your help btw

Also, how do i begin this question -
$(z +1)^4 + (z-1)^4 = 0$
• Jan 19th 2008, 04:04 PM
ThePerfectHacker
The first step is to find, $\sqrt[4]{1+i\sqrt{3}} = |1+i\sqrt{3}| \exp (i \arg (1+i\sqrt{3})/4 )$
Which is,
$\sqrt{1+3} \cdot e^{i\pi/ 12} = 2 \cos \frac{\pi}{12} + i 2\sin \frac{\pi}{12} = \left( 1 + \cos \frac{\pi}{6} \right) + i \left( 1 - \cos \frac{\pi}{6}\right)$
Which is,
$a = \left( 1 + \frac{\sqrt{3}}{2} \right) + i \left( 1 - \frac{\sqrt{3}}{2} \right)$
• Jan 19th 2008, 04:46 PM
topsquark
Quote:

Originally Posted by brd_7
and so how do i show it in terms of x+iy?

Thanks for all your help btw

Also, how do i begin this question -

$(z +1)^4 + (z-1)^4 = 0$

Tsk! New questions go in new threads.

I would expand:
$(z^4 + 4z^3 + 6z^2 + 4z + 1) + (z^4 - 4z^3 + 6z^2 - 4z + 1) = 0$

$2z^4 + 12z^2 + 2 = 0$
which is a biquadratic and easy to solve.

-Dan
• Jan 19th 2008, 06:50 PM
mr fantastic
Quote:

Originally Posted by topsquark
I favor putting the complex number into polar form:
$1 + i \sqrt{3} = re^{i \theta}$

Then go from there.

-Dan

I have a preference cutting straight to the chase and using $1 + i \sqrt{3} = re^{i (\theta + 2 n \pi)}, \, n \in \mathbb{Z}$.
• Jan 19th 2008, 07:11 PM
topsquark
Quote:

Originally Posted by mr fantastic
I have a preference cutting straight to the chase and using $1 + i \sqrt{3} = re^{i (\theta + 2 n \pi)}, \, n \in \mathbb{Z}$.

Well... (Sweating) If I want to do it the hard way, then let me! (Thinking)

-Dan
• Jan 19th 2008, 07:35 PM
ThePerfectHacker
I remember when I first learned de Moivre's theorem I did not like solving the equation $x^n = a$ because of the setting up that is involved. But here is my perferred method which turns this problem into an automatic problem.
• Jan 20th 2008, 03:25 AM
brd_7
The two roots that i gained were $+- \frac{12i}{2} + \frac{\sqrt(8)sqrt2}{2}$