### Author Topic: TUT5101  (Read 2010 times)

#### Yiheng Bian

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• Posts: 29
• Karma: 12
##### TUT5101
« on: January 26, 2020, 01:14:20 AM »
Find all solutions of the given equation:
$$z^8 =1$$
We know
$$z=r(\cos\theta +I\sin\theta)$$
$$z^8=r^8(\cos8\theta +I\sin8\theta)$$
Then
$$1=1+0i=1(\cos2k\pi + I\sin2k\pi)$$
So we have
$$r^8(\cos8\theta +I\sin8\theta) = 1(\cos2k\pi + I\sin2k\pi)$$
Therefore we get
$$r=1$$
$$8\theta=2k\pi$$
$$\theta= \frac{k\pi}{4}$$
So when
$$k=0, \theta=0,z=1(\cos0+i\sin0)=1$$
$$k=1, \theta=\frac{\pi}{4},z=1(\cos(\frac{\pi}{4})+I\sin{\frac{\pi}{4}})$$
$$k=2, \theta=\frac{\pi}{2},z=1(\cos(\frac{\pi}{2})+i\sin{\frac{\pi}{2}})$$
$$k=3, \theta=\frac{3\pi}{4},z=1(\cos(\frac{3\pi}{4})+i\sin{\frac{3\pi}{4}})$$
$$k=4, \theta=\pi,z=1(\cos\pi+i\sin\pi)$$
$$k=5, \theta=\frac{5\pi}{4},z=1(\cos(\frac{5\pi}{4})+i\sin{\frac{5\pi}{4}})$$
$$k=6, \theta=\frac{3\pi}{2},z=1(\cos(\frac{3\pi}{2})+i\sin{\frac{3\pi}{2}})$$
$$k=7, \theta=\frac{7\pi}{4},z=1(\cos(\frac{7\pi}{4})+i\sin{\frac{7\pi}{4}})$$
« Last Edit: January 26, 2020, 04:23:44 PM by Yiheng Bian »

#### Victor Ivrii

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##### Re: TUT5101
« Reply #1 on: January 26, 2020, 04:10:50 AM »
Escape cos, sin, log, .... : \cos, \sin, \log

#### Yiheng Bian

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• Posts: 29
• Karma: 12
##### Re: TUT5101
« Reply #2 on: January 26, 2020, 04:25:25 PM »
I have modified it sir.