phamminhphuong
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- Jan 15, 2016
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[FONT=KaTeX_Main][FONT=KaTeX_Math]#update 1: Thanks [/FONT][/FONT]ksdhart for your solution. I got what you mean. But I got another problem. The way the explain so hard to understand (khanacademy)
[FONT=KaTeX_Main][FONT=KaTeX_Math]Here is the problem :
Ex Z^3 = 27
The number 27 has a modulus of 27. The argument of 27 can be any multiple of 360 Decrees. So we can write k.360 for an integer k :
r^3 [ cos(3*O)+ i sin([/FONT][/FONT][FONT=KaTeX_Math]3*[/FONT]O)] = 27 [cos (k*360) + isin(k*360)] ===> 3*O = k*360[FONT=KaTeX_Main][FONT=KaTeX_Math]
From the first time they say It is strictly in 90 and 180. So i believe it should be 3*O = 90+ 180k.
Can you explain for me , please.
Z^5 = -7776i
[/FONT][/FONT]Find the solution of the following equation whose argument is strictly between 270, degree and [FONT=KaTeX_Main]360, degree[/FONT].[FONT=KaTeX_Main][FONT=KaTeX_Math]
Z = A + Bi
[/FONT][/FONT]
Follow the instruction, I know how to calculate the r.
This is how I calculate O:
5 * O = 270 decrees + k*360 decrees
O = 54 decrees + k*72 decrees
But about O. This is what they said :"Remember that O is strictly between [FONT=KaTeX_Main]270[/FONT] decrees and 360 decrees.Therefore, we need to find the multiple of [FONT=KaTeX_Main]90 decrees that is strictly within the range of [FONT=KaTeX_Main]270 decrees -54 decrees =216 decrees & 360-54=306 (decrees)
[/FONT][/FONT]. This multiple is simply 288 decrees. So O = 342 decrees.
Please help me, I don't understand how could they got 388 and 342 (decrees) [FONT=KaTeX_Main]
[/FONT]
[FONT=KaTeX_Main][FONT=KaTeX_Math]Here is the problem :
Ex Z^3 = 27
The number 27 has a modulus of 27. The argument of 27 can be any multiple of 360 Decrees. So we can write k.360 for an integer k :
r^3 [ cos(3*O)+ i sin([/FONT][/FONT][FONT=KaTeX_Math]3*[/FONT]O)] = 27 [cos (k*360) + isin(k*360)] ===> 3*O = k*360[FONT=KaTeX_Main][FONT=KaTeX_Math]
From the first time they say It is strictly in 90 and 180. So i believe it should be 3*O = 90+ 180k.
Can you explain for me , please.
Z^5 = -7776i
[/FONT][/FONT]Find the solution of the following equation whose argument is strictly between 270, degree and [FONT=KaTeX_Main]360, degree[/FONT].[FONT=KaTeX_Main][FONT=KaTeX_Math]
Z = A + Bi
[/FONT][/FONT]
Follow the instruction, I know how to calculate the r.
This is how I calculate O:
5 * O = 270 decrees + k*360 decrees
O = 54 decrees + k*72 decrees
But about O. This is what they said :"Remember that O is strictly between [FONT=KaTeX_Main]270[/FONT] decrees and 360 decrees.Therefore, we need to find the multiple of [FONT=KaTeX_Main]90 decrees that is strictly within the range of [FONT=KaTeX_Main]270 decrees -54 decrees =216 decrees & 360-54=306 (decrees)
[/FONT][/FONT]. This multiple is simply 288 decrees. So O = 342 decrees.
Please help me, I don't understand how could they got 388 and 342 (decrees) [FONT=KaTeX_Main]
[/FONT]
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