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# System prepended metadata

title: Wk1 Solution

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title: Wk1 Solution
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### Group 5
* **Q:** Given: $e^{ix} = cosx+isinx$
$P(x) = cos(x)+cos(2x)+cos(3x)+...+cos(100x)$
$P(4π/199) + 1/2 = cos(Aπ)$ 求A?

* **A:** $P(x) = \dfrac{e^{ix}+e^{-ix}}{2} + \dfrac{e^{i2x}+e^{-i2x}}{2} + ...+ \dfrac{e^{i100x}+e^{-i100x}}{2}$ 
= $\dfrac{e^{ix}(e^{i100x}-1)}{2(e^{ix}-1)} + \dfrac{e^{-ix}(e^{-i100x}-1)}{2(e^{-ix}-1)}$ 
= $\dfrac{e^{ix/2}(e^{i100x}-1)}{2(e^{ix/2}-e^{-ix/2})}$ + $\dfrac{e^{-ix/2}(e^{-i100x}-1)}{2(e^{-ix/2}-e^{ix/2})}$ 
= $\dfrac{e^{ix/2}e^{i100x}-e^{ix/2}-e^{-ix/2}e^{-i100x}+e^{-ix/2}}{2(e^{ix/2} - e^{-ix/2})}$ 
= $\dfrac{e^{i100x/2}-e^{-i100x/2}}{e^{ix/2}-e^{-ix/2}}{\dfrac{(e^{i101x/2}+e^{-i101x/2})}{2}}$ 
= $\dfrac{sin(100x/2)}{sin(x/2)}cos(101x/2)$ ($x$代入$\dfrac{4\pi}{199}$)
= $\dfrac{sin(\dfrac{\pi}{199})}{sin(\dfrac{2\pi}{199})}cos(\dfrac{3\pi}{199})$
= $\dfrac{sin(\dfrac{\pi}{199})}{2sin(\dfrac{\pi}{199})cos(\dfrac{\pi}{199})}(4cos^3(\dfrac{\pi}{199})-3cos(\dfrac{\pi}{199}))$

化簡可得 $A$ = $\dfrac{2}{199}$
助教： OK

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### Group 1

* **Q:** Let $z$ = $sin{\theta}$ - $icos{\theta}$, $0 < {\theta} <= \dfrac{\pi}{2}$ , find the principal argument of $z^2$.

* **A:** Let $u$ = $iz$ = $cos{\theta}$ + $isin{\theta}$. $z^{2}$ = - $u^{2}$, then $Arg($$z^2$$)$ = $Arg(-u^2)$ = $2{\theta} - \pi$
助教:OK

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### Group 2
* **Q:** 當$x→∞$，試求$(\frac{x}{\sqrt{2}}+\frac{x}{\sqrt{2}}i)的x次方根之所有解所圍成多邊形的面積。$

* **A:** ![](https://i.imgur.com/AYkdPZM.jpg)
助教:OK


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### Group 6
* **Q:**  z1=4(sqrt(3)+i)，其六次方根分別為w_0~w_5
且Arg(w_i)<Arg(w_(i+1)),i=0,1,2,3,4,5
求w1/w0+w3/w2+w5/w4為何？

* **A:** $|w_0| = |w_1| = |w_2| = |w_3| = |w_4| = |w_5|$ $arg(w_1) = arg(w_0) + \dfrac{\pi}{3}$,  $arg(w_3) = arg(w_2) + \dfrac{\pi}{3}$, $arg(w_5) = arg(w_4) + \dfrac{\pi}{3}$
$\dfrac{w_1}{w_0} + \dfrac{w_3}{w_2} + \dfrac{w_5}{w_4} = \dfrac{3+3\sqrt3 i}{2}$
助教:OK

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### Group 9

**Q:** 設z = cos72°+isin72°，透過z求出cos72°的值
(提示1: z^5 = 1)
(提示2:可以設 w =z+1/z 來先求出w的值) 

* **A:** ![](https://i.imgur.com/LuhEoSS.jpg)
助教:OK


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### Group 12
* **Q:** Let $k \in \mathbb{R}$. Prove that $|z + k|^{2} = (z + k)(\bar{z} + k)$, then express $|z + 1|^{2}$ as a function of $|z|$ and $Re(z)$ for all $z \in \mathbb{C}$.

* **A:**![](https://i.imgur.com/X4iM84K.jpg)
助教:OK


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### Group 13
* **Q:** Solve ${\left(\dfrac{1+\cos{1°}+\cos{2°}+i(\sin{1°}+\sin{2°})}{1+2\cos{1°}}\right)}^{90}$.

* **A:** ![](https://i.imgur.com/pfoVERN.jpg)
助教:OK

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### Group 15
>**Q** 設Z為一複數， |Z| = 1， theta = pi/30， 
>則 (1+cos(theta)+isin(theta))^5/((1+cos(theta)- isin(theta))^5 = 多少？

* **A:** because $|1+cos{\theta}+isin{\theta}| = |1+cos{\theta}-isin{\theta}|$ and $arg(1+cos{\theta}+isin{\theta})$ = $\dfrac{\pi}{60}$
$\dfrac{(1+cos{\theta}+isin{\theta})^5}{(1+cos{\theta}-isin{\theta})^5}$ = $(1\cdot e^{\dfrac{i\pi}{30}})^5$ = $1\cdot e^\dfrac{i\pi}{6}$ = $\dfrac{\sqrt3+i}{2}$
助教:OK

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### Group 16
考慮一個方程式$x^{4}=\alpha$的四個根，其中alpha為一複數,abs(alpha)=1，並且再令一複數z為這四根的乘積。請問：
(1)當z為純虛數時(實部為0)，四個根的主幅角各為多少？(答案不只一組)
(2)當z與其中一根相等時，求z的所有可能。

* **A:** ![](https://i.imgur.com/hkCPiXS.jpg)
助教:OK

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### Group 18
$z=x+yi,  x^{2}+y^{2}=4$(x,y are real numbers)
$Im(z^{4}-2z^{2}+1-2\cdot4^{2}z^{-2}+4^{4}z^{-4})=?$

* **A:** $z = 2e^{i\theta}$
 $Im(z^{4}-2z^{2}+1-2\cdot4^{2}z^{-2}+4^{4}z^{-4})$ =  $Im(16e^{4i\theta} -8e^{2i\theta} + 1  -8e^{-2i\theta}+ 16e^{-4i\theta})$ = $0$
 助教:OK

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### Group 19
>未知圖騰（日文︰アンノーン，英文︰Unown）是超能力屬性寶可夢。
>
>![](https://i.imgur.com/8cz3aiJ.png)
> 未知圖騰被稱為象徵寶可夢，它們的外表是黑色薄型的、如同文字，共有28種不同的型態，分別對應26個英文字母，問號和驚嘆號。評論家對未知圖騰的評價大多是負面的。

> 現在在電機系館發現了一個新型態的未知圖騰，該圖騰只生存於複數平面上，似乎擁有足以影響<b>系上同學成績</b>的力量。
> 該圖騰如下:
> 
> ![](https://imgur.com/p6ZuLbi.png)
> 
> 經研究發現其每天都會長1單位的頭髮，且每次生長的頭髮都與上個生長點與眼睛(原點)的連線垂直。(第一根頭髮便是眼睛右方Y方向之線段，<b>眼睛半徑為1</b>)
> 這個圖騰想剪頭髮，因為定期剪頭髮，他心情就會變好，心情好就會變成期中好運圖騰。你必須求出在第n天的時候，他的頭髮末端在哪，這樣你才能A+!
> 請幫他找到其座標，並以複數座標表示。

* **A:**![](https://i.imgur.com/JJWJfez.jpg)
助教:第一行$\sqrt{n+1}$...的形式是正確的，接著算出$\theta$們為多少即可，第二行以後是錯的

