z^6 = -64 URGENT MATH question HELP (1 Viewer)

Mr_Kap · May 25, 2016

how would i find the complex roots in z^6=-64.

i know it involves seomthing like r^6(cos(theta)+isin(theta)...and something like k=0,1,2,3,4,5

but i cant remember..

URGENT!!!! need help!!!

InteGrand · May 25, 2016

Mr_Kap said:
how would i find the complex roots in z^6=-64.

i know it involves seomthing like r^6(cos(theta)+isin(theta)...and something like k=0,1,2,3,4,5

but i cant remember..

URGENT!!!! need help!!!

$\noindent Note that $2^6 = 64$ and $-64$ = $64 e^{i \left(2k+1\right)\pi}$ for integers $k$. So the six roots will be $2e^{i\cdot \frac{\left(2k+1\right)\pi}{6}}$, for $k=0,1,\ldots,5$.$

Flop21 · May 25, 2016

Do you not have the mathsoc solutions??

-

Put it in polar form, like this:

z^6 = 64e^(i*Pi+2*k*Pi*i)

Then take each side to the power of 1/6:

z = [64e^(i*Pi+2*k*Pi*i)]^1/6

z = 2e^(i*Pi+2*k*Pi*i)/6

Now we sub in any values into k, say 0..5. (Can someone clarify if we just pick 6 (n) values here?).

to get the roots.

-

Now the second part of this question is usually, factorise into real linear and real irreducible quadratic factors right, so...

So find use the principle arguments only (between -Pi and Pi).

Now grab the conjugate pairs of the factors (Factors of p(z) = (z-root)). The conjugate pairs are the ones which are negatives of each other.

(z-2e^(Pi/6*I)) * (z-2e^-(Pi/6*I)) is the first one, do for the other 2.

Then expand this.

You'll get z^2 - 4([e^(Pi/6*I)+e^(Pi/6*I)]/2)z + 4

The part in brackets with the e's, is cos.

So it's cos(Pi/6) = root(3)/2

So z^2-2*root(3)z + 4 is one of the answers.

InteGrand · May 25, 2016

Flop21 said:
Now we sub in any values into k, say 0..5. (Can someone clarify if we just pick 6 (n) values here?).

You can pick any 6 (or generally n) consecutive numbers. Or more generally, any n numbers which have distinct values modulo n, i.e. distinct remainders when divided by n (but in practice, it's easiest to pick consecutive ones of course).

Flop21 · May 25, 2016

Flop21 said:
Then expand this.

You'll get z^2 - 4([e^(Pi/6*I)+e^(Pi/6*I)]/2)z + 4

The part in brackets with the e's, is cos.

So it's cos(Pi/6) = root(3)/2

So z^2-2*root(3)z + 4 is one of the answers.

Where does the 4 come from? And the /2?

Mr_Kap · May 25, 2016

guys guys...CIS FORM PLEASE!!!! please.. plzz...plz.zz.

havent learnt e^ipi yet

InteGrand · May 25, 2016

Flop21 said:
Where does the 4 come from? And the /2?

$\noindent It is due to expanding $\left(z-2\omega\right)\left(z-2\overline{\omega}\right)$, where for convenience we are writing $\omega = e^{i\frac{\pi}{6}}$, so note that $e^{-i\frac{\pi}{6}}=\overline{\omega}$. So (noting $\omega \overline{\omega}= \left |\omega\right |^2 = 1$), we have$

$\begin{align*}\left(z-2\omega\right)\left(z-2\overline{\omega}\right) &= z^2 - \left(2\omega +2\overline{\omega} \right)z + 2\omega\times 2\overline{\omega} \\ &= z^2 - 2\left(e^{i\frac{\pi}{6}}+e^{-i\frac{\pi}{6}}\right)z + 4.\end{align*}$

$\noindent If we note that $e^{i\frac{\pi}{6}}+e^{-i\frac{\pi}{6}}=2\Re \left(e^{i\frac{\pi}{6}}\right) = 2 \times \cos \frac{\pi}{6} = \sqrt{3}$, we have get a final expression of $z^2 - 2\sqrt{3}z+4$.$

$\noindent In general, $\left(z-\alpha\right)\left(z-\overline{\alpha}\right)=z^2 - 2\Re \left(\alpha\right)z + \left |\alpha\right |^2$ for any $z,\alpha \in \mathbb{C}$.$

InteGrand · May 25, 2016

Mr_Kap said:
guys guys...cis form please.. plzz...plz.zz.

havent learnt e^ipi yet

They are in fact the same thing: cis(t) = e^(it).

Mr_Kap · May 25, 2016

InteGrand said:
They are in fact the same thing: cis(t) = e^(it).

i still don't get how to do the initial question i asked..the e^i*Pi is confusing me. i distinctlly remmember doing this sort of question last year b4 i dropped but i cant find my notes

Flop21 · May 25, 2016

Mr_Kap said:
guys guys...cis form please.. plzz...plz.zz.

havent learnt e^ipi yet

???

What is cis, and what do you mean, I see this form in chap 3, page 111.

If your having issues just watch the revision video they made.

InteGrand · May 25, 2016

Mr_Kap said:
i still don't get how to do the initial question i asked..the e^i*Pi is confusing me. i distinctlly remmember doing this sort of question last year b4 i dropped but i cant find my notes

$\noindent Here it is with cis notation. Note $-64 = 64 \, \mathrm{cis} \left( \left(2k+1 \right) \pi \right)$, for $k$ integer. So the equation we have is $z^6 = 64\, \mathrm{cis}\left( \left(2k+1\right) \pi\right) \Rightarrow z = 2 \, \mathrm{cis}\left( \frac{\left( 2k+1 \right) \pi}{6}\right), \, k=0,1,\ldots, 5$.$

InteGrand · May 25, 2016

Flop21 said:
???

What is cis, and what do you mean, I see this form in chap 3, page 111.

If your having issues just watch the revision video they made.

The notation "cis" is commonly used in HSC 4U complex numbers (short for cos + i.sin. So cis(theta) is used to mean cos(theta) + i.sin(theta).).

Mr_Kap · May 25, 2016

Flop21 said:
???

What is cis, and what do you mean, I see this form in chap 3, page 111.

If your having issues just watch the revision video they made.

i dont have time to learn euler's bullshit. I'm behind in math...still at projections...so im relying on my 4u knowledge from last year to do complex numbers and shit.... meaning i need it in r(cos(theta)+i*sin(theta)) form.

Mr_Kap · May 25, 2016

InteGrand said:
$\noindent Here it is with cis notation. Note $-64 = 64 \text{cis}\left(\left(2k+1\right)\pi\right)$, for $k$ integer. So the equation we have is $z^6 = 64 \text{cis}\left(\left(2k+1\right)\pi\right)\Rightarrow z = 2 \text{cis}\left(\frac{\left(2k+1\right)\pi}{6}\right), \, k=0,1,\ldots, 5$.$

mistake somewhere in your latex

Flop21 · May 25, 2016

Mr_Kap said:
i dont have time to learn euler's bullshit. I'm behind in math...still at projections...so im relying on my 4u knowledge from last year to do complex numbers and shit.... meaning i need it in r(cos(theta)+i*sin(theta)) form.

Well really all you have to remember is:

e^(i*Pi + 2k*Pi*i)

and shove the modulus to the left, and bam do what I said to solve, you're just using basic algebra and knowledge of exponentials/powers from there.

Trust me I don't really understand what's going on but it's what you do for the question. I feel like changing it into cis form would make it more complicated.

But hopefully integrand or someone can help you out.

Mr_Kap · May 25, 2016

InteGrand said:
$\noindent Here it is with cis notation. Note $-64 = 64 \, \mathrm{cis} \left( \left(2k+1 \right) \pi \right)$, for $k$ integer. So the equation we have is $z^6 = 64\, \mathrm{cis}\left( \left(2k+1\right) \pi\right) \Rightarrow z = 2 \, \mathrm{cis}\left( \frac{\left( 2k+1 \right) \pi}{6}\right), \, k=0,1,\ldots, 5$.$

legend brother

InteGrand · May 25, 2016

Flop21 said:
Well really all you have to remember is:

e^(i*Pi + 2k*Pi*i)

and shove the modulus to the left, and bam do what I said to solve, you're just using basic algebra and knowledge of exponentials/powers from there.

Trust me I don't really understand what's going on but it's what you do for the question. I feel like changing it into cis form would make it more complicated.

But hopefully integrand or someone can help you out.

I think the reason he asked for cis notation is that he's (at least for now) more used to that than exponential form since he remembers it from 4U (whereas exponential form isn't in 4U).

anomalousdecay · May 26, 2016

From the words of Doust:

Doust said:
Don't use cis notation. It's an abomination. Expand it out to cos(x) + isin(x) or use exponential form.

leehuan · May 26, 2016

Lol yep agreed. Cut it out with the cis already.

@Flop pretend cis never existed in your life lol. It only (still) exists for MX2 and IB because it's probably a bit early to show that it's actually e^(i.theta)

z^6 = -64 URGENT MATH question HELP (1 Viewer)

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