Quod Erat Demonstrandum

2011/11/18

polar form

Filed under: Pure Mathematics — johnmayhk @ 11:46 上午
Tags: ,

基礎題:

Convert z=-\cos \theta -i\sin\theta into polar form.

問:「在何象限,cosine 和 sine 值也是負?」 (這是極不精確的說法,但學生又明我說甚麼。)

答:「第三。」

回:「好!那如何把角度 adjust 到第三象限?」

答:「加 \pi。」

回:「對!那麼 -\cos \theta 就可用 \cos(\pi +\theta) 代替,而 -\sin \theta 就用 \sin(\pi +\theta) 代之。故 -\cos \theta -i\sin\theta 的 polar form 是 \cos(\pi +\theta)+i\sin(\pi +\theta) 也。」

利用"adjust"角度到不同象限的想法,不難有:

-\cos \theta +i\sin\theta =\cos (\pi-\theta) +i\sin (\pi-\theta)

\cos \theta -i\sin\theta =\cos (2\pi-\theta) +i\sin (2\pi-\theta)=\cos(-\theta) +i\sin(-\theta)

進一步問,諸如

Find the principal value of the argument of -\cos (\frac{5\pi}{6}) -i\sin(\frac{5\pi}{6}).

先轉之為 polar form,即

z=-\cos (\frac{5\pi}{6}) -i\sin(\frac{5\pi}{6})

=\cos (\pi+\frac{5\pi}{6}) +i\sin(\pi+\frac{5\pi}{6})

=\cos (\pi+\frac{5\pi}{6}-2\pi) +i\sin(\pi+\frac{5\pi}{6}-2\pi) [因 principle value of argument Arg(z) 要規限為 -\pi < Arg(z) \le \pi。]

Arg(z)=\pi+\frac{5\pi}{6}-2\pi=-\frac{\pi}{6}

注:使用大寫的 Arg(z),希望使它有別於多值函數 arg(z)。如上例,Arg(z)=-\frac{\pi}{6},而 arg(z)=-\frac{\pi}{6}+2n\pi(where n \in \mathbb{Z}

習題:

Convert the following into polar form.

1. \sin\theta+i\cos\theta

2. \sin\theta-i\cos\theta

3. -\sin\theta+i\cos\theta

4. -\sin\theta-i\cos\theta

Answers:

1. \cos(\frac{\pi}{2}-\theta)+i\sin(\frac{\pi}{2}-\theta)

2. \cos(\frac{3\pi}{2}+\theta)+i\sin(\frac{3\pi}{2}+\theta)

3. \cos(\frac{\pi}{2}+\theta)+i\sin(\frac{\pi}{2}+\theta)

4. \cos(\frac{3\pi}{2}-\theta)+i\sin(\frac{3\pi}{2}-\theta)

1 則迴響 »

  1. 一理通馬國明

    迴響 由 FTL — 2011/11/20 @ 9:41 上午 | 回覆


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