Quod Erat Demonstrandum

2011/11/18

polar form

Filed under: Pure Mathematics — johnmayhk @ 11:46 上午
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Convert $z=-\cos \theta -i\sin\theta$ into polar form.

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

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

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

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

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

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

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

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. $-\sin\theta-i\cos\theta$

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 | 回應