Author Topic: TUT5101  (Read 1105 times)

Yiheng Bian

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