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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: darkphoenix on March 07, 2010, 09:03:45 pm

Title: Complex Numbers Help
Post by: darkphoenix on March 07, 2010, 09:03:45 pm: Got a question, couldn't quite figure it out.

1. Write down a polynomial of degree 3, whose coefficients are all real, that has 4i and 2 as two of its zeros.

I tried using simultaneous equations, but it didn't quite work.
Title: Re: Complex Numbers Help
Post by: mandy on March 07, 2010, 09:06:04 pm: Seeing as the coefficients are all real, wouldn't you use the Conjugate Root Theorem? Then, you would have three factors that you could expand and get the polynomial from.

i.e. $(z - 4i)(z + 4i)(z - 2)$

...I hope that's right.
Title: Re: Complex Numbers Help
Post by: Martoman on March 07, 2010, 09:07:32 pm: Quote from: mandy on March 07, 2010, 09:06:04 pm
Seeing as the coefficients are all real, wouldn't you use the Conjugate Root Theorem? Then, you would have three factors that you could expand and get the polynomial from.

i.e. $(z - 4i)(z + 4i)(z - 2)$

...I hope that's right.

You are correct. You beat me to it :)
Title: Re: Complex Numbers Help
Post by: darkphoenix on March 07, 2010, 09:24:42 pm: Omg LOL

Thanks alot mandy. :)

Dam i didnt think it was that simple and i started subbing stuff in..ah never mind.
Title: Re: Complex Numbers Help
Post by: mandy on March 07, 2010, 09:31:21 pm: Quote from: darkphoenix on March 07, 2010, 09:24:42 pm
Omg LOL

Thanks alot mandy. :)

Dam i didnt think it was that simple and i started subbing stuff in..ah never mind.

Haha, glad I could help :)
Title: Re: Complex Numbers Help
Post by: darkphoenix on March 07, 2010, 11:46:24 pm: Ok i have another question:

2. Factorise $z^3 + i$ over C
Title: Re: Complex Numbers Help
Post by: kamil9876 on March 08, 2010, 12:00:21 am: solve $z^3+i=0$

then the polynomial is $(z-z_1)(z-z_2)(z-z_3)$ where $z_1,z_2,z_3$ are the roots.
Title: Re: Complex Numbers Help
Post by: Martoman on March 08, 2010, 12:05:28 am: Quote from: kamil9876 on March 08, 2010, 12:00:21 am
solve $z^3+i=0$

then the polynomial is $(z-z_1)(z-z_2)(z-z_3)$ where $z_1,z_2,z_3$ are the roots.

Alternatively

$z^{3} + ((i)^{\frac{1}{3}})^3$

Then simply factorise this as a perfect cube. However, this would still facilitate solving for $(i)^{\frac{1}{3}}$ , essentially what Kamil is talking about.
Title: Re: Complex Numbers Help
Post by: darkphoenix on March 08, 2010, 12:14:00 am: Hm ive managed to get $z - i$ as one of the factors but cant seem to find the other quadratic factor
Title: Re: Complex Numbers Help
Post by: Martoman on March 08, 2010, 12:21:28 am: Use demoivre's theorm.

$z^3 = -i$

Then $r^3cis(3\theta) = cis(\frac{-\pi}{2} + 2\pi k) \; where \; k\epsilon Z$

Solutions occur by equating sides, so $r^3 = 1, <br />3\theta = \frac{-\pi}{2} + 2\pi k$

Then $\theta = \frac{1}{3} (-\frac{\pi}{2} + 2\pi k)$

Now solutions occur at: $cis(\frac{1}{3} (-\frac{\pi}{2} + 2\pi k))$
Shove in values of k until it repeats. I would shove in $k=-1,0,1$ Once you done that convert to cartesian. Then you have your roots.
Title: Re: Complex Numbers Help
Post by: Martoman on March 08, 2010, 12:22:54 am: Quote from: darkphoenix on March 08, 2010, 12:14:00 am
Hm ive managed to get $z - i$ as one of the factors but cant seem to find the other quadratic factor

Alternatively $\frac{z^3 + i}{z-i}$ then just divide however you do, normally or synthetically.
Title: Re: Complex Numbers Help
Post by: darkphoenix on March 08, 2010, 12:44:59 am: Alright got it.

Cheers dude.