Sorry... maths help again.

Mason-

Mason-

Mason-

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fail... can someone move this to general discussion.

Thank you whoever moved this for me ^^


z^4 = 16 get the roots in cartesian form.

So I worked them out in polar form easy. cos(2pi)k + isin(2pi)k
where k is 1-3

Normally putting them back into cartesian form is easy but I am unsure about how the = 16 part works.

I've only done it where it = 1.

do I just change it to:

1/16(cos(2pi)k + isin(2pi)k) = 1

and use magnitude 1/16

Also the next part is annoying me just by looking at it.

solve

w^4 = 16(1-w)^4

again in cartesian form

I don't want you to tell me the answer but if you could give me some pointers. thanks.
 
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I was reading gfx card forum.
Left PC,

I'm usually reading the general discussion form and so just started a thread only noticed after I posted it. Needless to say, facepalm.
 
Mason- said:
z^4 = 16 get the roots in cartesian form.

So I worked them out in polar form easy. cos(2pi)k + isin(2pi)k
where k is 1-3
Click to expand...

Try a substitution z=2v, then v^4 = 1. In polar form we have v = e^(2*pi*k*i/4), where k is 1-4. So then we get z = 2e^(2*pi*k*i/4).

To get them in cartesian form, just use Euler i.e. e^(i*theta) = cos(theta) + i sin(theta)
 
Ok, so we want the 4 roots of z^4 = 16. cool.


First we need magnitude in the complex plane. Simply, |z| = 4th root of 16 = 2.

Now, we need direction in the complex plane. We know from roots of unity that they'll lie equiangular..er..with the same angular intervals between them. What does that mean? It means that each z will lie on an axis, as that breaks the complex plane up into 4 equal parts. So, we can write it as

z = r*e^{i*(j*pi)/4}. This will break the complex plane up into 4 quarters as we vary j, with each z lying on a horizontal/vertical line if you visualise it.

We know r = 2, so we just have the 4 roots for j = 1, 2, 3 and 4.

Rewrite z in the polar form: z = r*cos(..) + i*r*sin(..) and job done!
 
Mason- said:
Also the next part is annoying me just by looking at it.

solve

w^4 = 16(1-w)^4

again in cartesian form

I don't want you to tell me the answer but if you could give me some pointers. thanks.
Click to expand...

You know how to solve the equation z^4 = 16 from doing the first bit. If you divide through by (1-w)^4 and subsitute u = w/(1-w) then you can solve for u and work out the corresponding w after.
 
Mason- said:
z^4 = 16 get the roots in cartesian form.

So I worked them out in polar form easy. cos(2pi)k + isin(2pi)k
where k is 1-3
Click to expand...
That isn't polar form. Polar form would be of the form (r,theta), where z = r e^{i.theta}. Using Euler's formula for e^{i.t} = cos(t) + i sin(t) you can then convert into Cartesians (x,y) by z = r cos(t) + i r sin(t) = x + i y, so x = r cos(t) and y = r sin(t).

As pointed out Using z = r e^{i.t} gives r^4 e^{4.i.t} = 16 so you end up solving r^4 = 16 and e^{4.i.t} = 1 = e^{2.pi.k} for k = 0,1,2,3.

Mason- said:
Also the next part is annoying me just by looking at it.

solve

w^4 = 16(1-w)^4

again in cartesian form
Click to expand...
Given the factor of 16 I would imagine the point is to rearrange that to make it look like the first part of the question (ie rearrange to get 16 = f(w)^4 ) and since you know how to solve 16 = z^4 you then use z = f(w), rearranging to give w = f^{-1}(z) and put in the expressions for z you know.

If you're unsure what I'm talking about just say.

/edit Beaten to it.
 
I've just worked this out the hard way.
Z^4 = 16

I can gather straight away that:
Z = 2, -1, a+ib, a-ib

(a+ib) ^ 4 = 16
=>
a^4 + b^4 -6a^2b^2 +a^4b^3i -a^3b^4i = 16 + 0i

=> a = b and a^4 = -4
=> a = i + 1 and b = i + 1

Thus the solution is ( i(i+1) + i + 1) ^ 4 = 16
Which simplifies to 2i and -2i
 
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