Author Topic: complex numbers question (Read 1530 times) Tweet Share

greedy · « **on:** March 27, 2011, 03:43:26 pm »

hi... I am having trouble expressing the solutions to polar form:

determine all the complex solutions of:

Z^4 - 2i z^2 + 1 = 0

answers needs to be in polar form with -pi < theta < pi

thx for the help...T T

enpassant · « **Reply #1 on:** March 27, 2011, 03:56:42 pm »

z=sqrt(1-sqrt2)cis(pi/4)
z=sqrt(1-sqrt2)cis(-3pi/4)
z=sqrt(1+sqrt2)cis(pi/4)
z=sqrt(1+sqrt2)cis(-3pi/4)

VCE247 · « **Reply #2 on:** March 27, 2011, 04:03:58 pm »

how did you get that enpassant
Z^4 - 2i z^2 + 1 = 0

Doesnt that =
(z^2 + i^2) ^2?

so
z^2 + i^2 = 0
z^2 = 1
then solve from there?

oops dw silly mistake

greedy · « **Reply #3 on:** March 27, 2011, 08:39:30 pm »

Quote from: enpassant on March 27, 2011, 03:56:42 pm

z=sqrt(1-sqrt2)cis(pi/4)
z=sqrt(1-sqrt2)cis(-3pi/4)
z=sqrt(1+sqrt2)cis(pi/4)
z=sqrt(1+sqrt2)cis(-3pi/4)

thats what calculater given me...

looks odd...

can u plz show 1 of the solution after u find z in catesian?

thx T T

luffy · « **Reply #4 on:** March 27, 2011, 09:06:17 pm »

Here is how I would do the question: (I hope I didn't make any mistakes - Sorry if I did)

$z^4 - 2iz^2 + 1 = 0$

Use the quadratic formula/complete the square to solve for z^2:
$(z^2 - i)^2 + 2 = 0$

$(z^2 - i + \sqrt{2}i)(z^2 - i - \sqrt{2}i) = 0$

$z^2 = i \pm \sqrt{2}i$

$z^2 = (1 \pm \sqrt{2})i$

Convert z^2 into polar form:
$Arg(z^2) = \frac{\pi}{2} \rightarrow \textup{ Because } Re(z) = 0 \textup {, } Im(z) = \textup{ non-zero constant}$

$|z^2| = (1 \pm \sqrt{2})$

Thus, in polar form:
$\therefore z^2 = (1 \pm \sqrt{2})cis(\frac{\pi}{2} + 2n\pi)$

Using De Moivre's Theorem:
$\therefore z = \sqrt{(1 \pm \sqrt{2})}cis(\frac{\pi}{4} + n\pi)$

As $-\pi < Arg(z) \leq \pi$
$z = \sqrt{(1 + \sqrt{2})}cis(\frac{\pi}{4})$
$z = \sqrt{(1 - \sqrt{2})}cis(\frac{\pi}{4})$
$z = \sqrt{(1 + \sqrt{2})}cis(\frac{-3\pi}{4})$
$z = \sqrt{(1 - \sqrt{2})}cis(\frac{-3\pi}{4})$

Hope I helped.

enpassant · « **Reply #5 on:** March 27, 2011, 09:12:20 pm »

Z^4 - 2i z^2 + 1 = 0
Z^4 - 2i z^2 - 1 = -2
(z^2-i)^2=-2
(z^2-i)^2=2cis(pi+2kpi)
z^2-i=sqrt2cis((pi+2kpi)/2)
z^2-i=isqrt2 or -isqrt2
z^2=(1+sqrt2)i or (1-sqrt2)i
z^2=(1+sqrt2)cis(pi/2+2kpi) or (1-sqrt2)cis(pi/2+2kpi)
etc

TrueTears · « **Reply #6 on:** March 27, 2011, 09:19:59 pm »

just do this, let u = z^2

rest is pretty simple.

greedy · « **Reply #7 on:** March 28, 2011, 03:46:38 am »

Quote from: luffy on March 27, 2011, 09:06:17 pm

Here is how I would do the question: (I hope I didn't make any mistakes - Sorry if I did)

$z^4 - 2iz^2 + 1 = 0$

Use the quadratic formula/complete the square to solve for z^2:
$(z^2 - i)^2 + 2 = 0$

$(z^2 - i + \sqrt{2}i)(z^2 - i - \sqrt{2}i) = 0$

$z^2 = i \pm \sqrt{2}i$

$z^2 = (1 \pm \sqrt{2})i$

Convert z^2 into polar form:
$Arg(z^2) = \frac{\pi}{2} \rightarrow \textup{ Because } Re(z) = 0 \textup {, } Im(z) = \textup{ non-zero constant}$

$|z^2| = (1 \pm \sqrt{2})$

Thus, in polar form:
$\therefore z^2 = (1 \pm \sqrt{2})cis(\frac{\pi}{2} + 2n\pi)$

Using De Moivre's Theorem:
$\therefore z = \sqrt{(1 \pm \sqrt{2})}cis(\frac{\pi}{4} + n\pi)$

As $-\pi < Arg(z) \leq \pi$
$z = \sqrt{(1 + \sqrt{2})}cis(\frac{\pi}{4})$
$z = \sqrt{(1 - \sqrt{2})}cis(\frac{\pi}{4})$
$z = \sqrt{(1 + \sqrt{2})}cis(\frac{-3\pi}{4})$
$z = \sqrt{(1 - \sqrt{2})}cis(\frac{-3\pi}{4})$

Hope I helped.

thats really well explained, I have learnt from it.

you helped me alot!!!

thank you so much!

PS: really clear layout, I will learn to do that as well

Search the forums now!

We have moved!

Author Topic: complex numbers question (Read 1530 times) Tweet Share

greedy

complex numbers question

enpassant

Re: complex numbers question

VCE247

Re: complex numbers question

greedy

Re: complex numbers question

luffy

Re: complex numbers question

enpassant

Re: complex numbers question

TrueTears

Re: complex numbers question

greedy

Re: complex numbers question

Recent Posts

User:
Password: