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MC III-B Help

 
 
SongKevin的头像
MC III-B Help
SongKevin - 2022年07月12日 Tuesday 15:35
 

Hi, I need help with 28, 29, 30,33,32, and 35

 
SongKevin的头像
Re: MC III-B Help
SongKevin - 2022年07月13日 Wednesday 15:38
 

I meant MC III, not III-B

WangDr. Kevin的头像
Re: MC III-B Help
WangDr. Kevin - 2022年07月14日 Thursday 01:40
 

I will give you some hints.  If you include some details on what you have done so far, it would be better.

2.28: Use the Factor Theorem.  Plug in $x=1$ and $x=$ the imaginary cube root of unity.

2.29: Expand everything and clear the denominator, and see where that leads you,

2.30: This polynomial looks similar to a binomial expansion of degree 4.

2.32: The two sums can be combined into one, using one of them as the real part and the other as the imaginary part to form a sum of complex numbers.

2.33: Consider the binomial expansion of $(1+x)^n$, then plug in $x=\omega$ where $\omega$ is the imaginary cube root of unity.  We have the property $\omega^3=1$ and $\omega^2 + \omega + 1=0$.

2.35(a): This product can be derived from expressing the polynomial of $z^{2n}-1$ as a product of linear factors, and each factor represent one of its roots. The part (b) is similar.  Keep in mind that the roots of unity can be paired up, each pair containing two conjugate complex numbers.

SongKevin的头像
MC III Help
SongKevin - 2022年07月15日 Friday 08:14
 

For 2.30, I have 

(x-i)4  = 1+i

            =  \( \sqrt{2} \cdot e^{ \pi i/4} \)

But I don't know where to go from there.


For 2.32, I have

C = sum of the cosines, and S = sum of the sines.

C + iS = \( \sum{k \cdot e^{2k \pi i/n}} \) from 1 to n-1 and I am stuck.


For 2.35, I can't see how you write it as a product of linear factors of \( z^{2n}-1 \). 

WangDr. Kevin的头像
Re: MC III Help
WangDr. Kevin - 2022年07月16日 Saturday 14:11
 

See my answer to the following post for some of the details:

https://classes.areteem.org/mod/forum/discuss.php?d=1193

I will add a few more steps in another reply.

WangDr. Kevin的头像
Re: MC III Help
WangDr. Kevin - 2022年07月16日 Saturday 14:23
 

2.30 talks about the area of a convex polygon whose vertices are the roots of the equation.  The equation has been written as a fourth power of a complex number equals another complex number.  So the four vertices actually form a square in the complex plane.  You need to figure out the area of the square.  So what matters is the length from the center to a vertex.  We can solve for the actual roots if we want to (there is a formula for that in the textbook).  But you can also figure out only the length.

2.32: you need to find out how to calculate the sum of the following type:

$$ X = 1 + 2r + 3r^2 + 4r^3 + \cdots + nr^{n-1} $$

Hint: Write out $rX$, and calculate $X - rX$, and see what you get.