Complex q (2 Viewers)

helpfulperson? · Feb 3, 2024

How would you do q2b. I’m having trouble trying to get cos4theta. Thanks!

Luukas.2 · Feb 3, 2024

helpfulperson? said:
How would you do q2b. I’m having trouble trying to get cos4theta. Thanks!

Some immediate thoughts:

What did you get for 2(a)?
Have you noticed the two pairs of differences of two squares on the RHS?
- And that the resulting $\cos^2{\alpha}$ terms can be found by rewriting into $\cos{2\alpha}$ terms?
$\cos{4\theta}\ \text{is the real part of}\ z^4\ \text{when}\ z = \cos{\theta} + i\sin{\theta}\ \text{after applying De Moivre's Theoerm}$

Lith_30 · Feb 3, 2024

idk if there is an easier way but...

the from part a you get the quadratic factors as
$\\z^8+1=(z^2+2z\cos\left(\frac{\pi}{8}\right)+1)(z^2-2z\cos\left(\frac{\pi}{8}\right)+1)(z^2+2z\cos\left(\frac{3\pi}{8}\right)+1)(z^2-2z\cos\left(\frac{3\pi}{8}\right)+1)$

now we use the substitution $z=e^{i\theta}$ ,

$\\e^{8i\theta}+1=(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)(e^{2i\theta}-2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{3\pi}{8}\right)+1)(e^{2i\theta}-2e^{i\theta}\cos\left(\frac{3\pi}{8}\right)+1)$

now lets just look at one factor of that expression, say $(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)$

expanding into the mod argument form...
$\cos(2\theta)+2\cos\left(\theta\right)\cos\left(\frac{\pi}{8}\right)+1+i(\sin(2\theta)+2\sin\left(\theta\right)\cos\left(\frac{\pi}{8}\right))\\\\=2\cos^2\left(\theta\right)+2\cos\left(\theta\right)\cos\left(\frac{\pi}{8}\right)+2i\sin\left(\theta\right)(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))\\\\=2(\cos\left(\theta\right)+i\sin\left(\theta\right))(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))$

now all the other factors can be simplified similarly to get
$\\16(\cos\left(\theta\right)+i\sin\left(\theta\right))^4(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))\\\\=16e^{4i\theta}(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))$

Now we include the LHS to simplify the expression

$\\e^{8i\theta}+1=16e^{4i\theta}(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))\\\\e^{4i\theta}+e^{-4i\theta}=16(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))\\\\\cos(4\theta)=8(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))$

as required

ExtremelyBoredUser · Feb 4, 2024

Lith_30 said:
idk if there is an easier way but...

the from part a you get the quadratic factors as
$\\z^8+1=(z^2+2z\cos\left(\frac{\pi}{8}\right)+1)(z^2-2z\cos\left(\frac{\pi}{8}\right)+1)(z^2+2z\cos\left(\frac{3\pi}{8}\right)+1)(z^2-2z\cos\left(\frac{3\pi}{8}\right)+1)$

now we use the substitution $z=e^{i\theta}$ ,

$\\e^{8i\theta}+1=(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)(e^{2i\theta}-2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{3\pi}{8}\right)+1)(e^{2i\theta}-2e^{i\theta}\cos\left(\frac{3\pi}{8}\right)+1)$

now lets just look at one factor of that expression, say $(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)$

expanding into the mod argument form...
$\cos(2\theta)+2\cos\left(\theta\right)\cos\left(\frac{\pi}{8}\right)+1+i(\sin(2\theta)+2\sin\left(\theta\right)\cos\left(\frac{\pi}{8}\right))\\\\=2\cos^2\left(\theta\right)+2\cos\left(\theta\right)\cos\left(\frac{\pi}{8}\right)+2i\sin\left(\theta\right)(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))\\\\=2(\cos\left(\theta\right)+i\sin\left(\theta\right))(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))$

now all the other factors can be simplified similarly to get
$\\16(\cos\left(\theta\right)+i\sin\left(\theta\right))^4(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))\\\\=16e^{4i\theta}(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))$

Now we include the LHS to simplify the expression

$\\e^{8i\theta}+1=16e^{4i\theta}(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))\\\\e^{4i\theta}+e^{-4i\theta}=16(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))\\\\\cos(4\theta)=8(\cos\left(\theta\right)+\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{\pi}{8}\right))(\cos\left(\theta\right)+\cos\left(\frac{3\pi}{8}\right))(\cos\left(\theta\right)-\cos\left(\frac{3\pi}{8}\right))$

as required

never gonna see these qs again, miss those days

Lith_30 · Feb 4, 2024

ExtremelyBoredUser said:
never gonna see these qs again, miss those days

same ngl, back when maths was comprehensible

also bruh for my solution just factor out $e^{i\theta}$ for each of these $(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)$ . waaaay simpler

ExtremelyBoredUser · Feb 4, 2024

Lith_30 said:
same ngl, back when maths was comprehensible

also bruh for my solution just factor out $e^{i\theta}$ for each of these $(e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1)$ . waaaay simpler

yea bro u did it the long way, much more easier just using the identity (e^ix + e^-ix) = 2cos(x) and then its more cleanerr

liamkk112 · Feb 4, 2024

ExtremelyBoredUser said:
never gonna see these qs again, miss those days

is unsw maths as brain toasting as others say

ExtremelyBoredUser · Feb 4, 2024

liamkk112 said:
is unsw maths as brain toasting as others say

first year no. second year idk but seems so.

wouldn't say its a walk in a park tho

ExtremelyBoredUser · Feb 4, 2024

oh but fk inclusion exclusion principle tho and the bs puzzle questions

ivanradoszyce · Feb 6, 2024

helpfulperson? said:
View attachment 42312
How would you do q2b. I’m having trouble trying to get cos4theta. Thanks!

Can you please tell me where you got this question. I like it a lot!

Aeonium · Feb 6, 2024

ivanradoszyce said:
Can you please tell me where you got this question. I like it a lot!

Kurt mathematics in Epping

ivanradoszyce · Feb 11, 2024

Attached is a complete solution written in Latex

tywebb · Feb 11, 2024

Here is a more efficient method to solve 3a.

$\text{If }T_n\text{ is the }n\text{-th Chebychev polynomial of the first kind so that}$
$T_n(\cos\theta)=\cos(n\theta)\text{ then using the }{}_2F_1\text{ hypergeometric we have that}$
$T_n(x)=\textstyle{}_2F_1(-n,n;\frac{1}{2};\frac{1-x}{2})$
$\text{and so }T_5(x)=\textstyle{}_2F_1(-5,5;\frac{1}{2};\frac{1-x}{2})=16x^5-20x^3+5x$
$\text{and since }T_{2n+1}(\sin\theta)=(-1)^n\sin((2n+1)\theta)\text{ we now have}$
$\begin{aligned}\tan(5\theta)&=\textstyle\frac{\sin5\theta}{\cos5\theta}\\ &=\textstyle\frac{T_5(\sin\theta)}{T_5(\cos\theta)}\\ &=\textstyle\frac{16\sin^5\theta-20\sin^3\theta+5\sin\theta}{16\cos^5\theta-20\cos^3\theta+5\cos\theta}\div\frac{\cos^5\theta}{\cos^5\theta}\\ &=\textstyle\frac{16\tan^5\theta-20\tan^3\theta\sec^2\theta+5\tan\theta\sec^4\theta}{16-20\sec^2\theta+5\sec^4\theta}\\ &=\textstyle\frac{16\tan^5\theta-20\tan^3\theta(1+\tan^2\theta)+5\tan\theta(1+\tan^2\theta)^2}{16-20(1+\tan^2\theta)+5(1+\tan^2\theta)^2}\\ &=\textstyle\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}\end{aligned}$

Luukas.2 · Feb 11, 2024

ivanradoszyce said:
Attached is a complete solution written in Latex

There are a couple of issues with question 1:

On page 2, you have

$|z| = \sqrt{4\cos^2{\frac{\alpha}{2}}} = 2\cos{\frac{\alpha}{2}}$

but this is not always true:

$\begin{align*} \text{Take $\alpha = 2\pi$:} \qquad \cos{\frac{\alpha}{2}} &= \cos{\pi} = -1 \\ |z| &= \sqrt{4\times(-1)^2} = 2 \\ \\ \text{But} \qquad 2\cos{\frac{\alpha}{2}} &= -2 \neq |z| \end{align*}$

The simplification only applies over restricted domains like $-\pi \le \alpha \le \pi$ , and the result should actually be

$|z| = \sqrt{4\cos^2{\frac{\alpha}{2}}} = 2\left|\cos{\frac{\alpha}{2}}\right|$

A similar problem arises from the following statement on the same page, that

$\text{Arg}\,z = \tan^{-1}{\frac{y}{x}} \text{, where $z = x + iy$}$

as this statement is undefined for $x = 0$ and incorrect if $x < 0$ . The inverse tangent function outputs angles covering only about half of the range of the principal argument.

$\text{For example, if $z = -1 - i$, then $\arg{z} = -\frac{3\pi}{4}$ but $\tan^{-1}{\frac{y}{x}} = \tan^{-1}{\frac{-1}{-1}} = \tan^{-1}{(1)} = \frac{\pi}{4} \neq \arg{z}$.}$

In effect, this proof establishes the result for some, but not all, possible values of $\alpha$ .

ivanradoszyce · Feb 11, 2024

Thank-you very much Lith_30. I'll edit those domain restrictions to page 2.

Luukas.2 · Feb 11, 2024

ivanradoszyce said:
Thank-you very much Lith_30. I'll edit those domain restrictions to page 2.

If you are going to make edits, might I suggest the cases where you put

$(1 + \cos\alpha + i\sin\alpha)^k + (1 + \cos\theta - i\sin\theta)^k$

on pages 1 and 3 have the $\theta$ 's changed to $\alpha$ ?

ivanradoszyce · Feb 11, 2024

Got it - thanks.

tywebb · Feb 11, 2024

Here is a more efficient solution to 1.

$\begin{aligned}(1+\cos\alpha+i\sin\alpha)^k+(1+\cos\alpha-i\sin\alpha)^k&=(1+e^{i\alpha})^k+(1+e^{-i\alpha})^k\\ &=\textstyle\sum_{j=0}^k{k\choose j}(e^{i\alpha j}+e^{-i\alpha j})\\ &=\textstyle\sum_{j=0}^k{k\choose j}(e^{i\alpha j}+e^{i\alpha(j-k)})\text{ by symmetry}\\ &=\textstyle\sum_{j=0}^k{k\choose j}(e^{\frac{ik\alpha}{2}+\frac{ij\alpha}{2}-\frac{i(k-j)\alpha}{2}}+e^{-\frac{ik\alpha}{2}+\frac{ij\alpha}{2}-\frac{i(k-j)\alpha}{2}})\\ &=\textstyle(e^\frac{ik\alpha}{2}+e^{-\frac{ik\alpha}{2}})\sum_{j=0}^k{k\choose j}e^\frac{ij\alpha}{2}\cdot e^{-\frac{i(k-j)\alpha}{2}}\\ &=(e^\frac{ik\alpha}{2}+e^{-\frac{ik\alpha}{2}})(e^\frac{i\alpha}{2}+e^{-\frac{i\alpha}{2}})^k\\ &=\textstyle2\cos\frac{k\alpha}{2}(2\cos\frac{\alpha}{2})^k\\ &=\textstyle2^{k+1}\cos\frac{k\alpha}{2}\cos^k\frac{\alpha}{2} \end{aligned}$

Lith_30 · Feb 11, 2024

ivanradoszyce said:
Thank-you very much Lith_30. I'll edit those domain restrictions to page 2.

What did I say???

ExtremelyBoredUser · Feb 11, 2024

Lith_30 said:
What did I say???

Sometimes its what you don't say.

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