Power of complex number

phamminhphuong

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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]
 

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I'm sorry, but the images are too small for me to be able to read. Kindly please reply with the typed-out text, or else clear images. Thank you! ;)
 
I have to be honest, I haven't any earthly idea what your web program is trying to explain there. It might as well be an alien language, as far as I'm concerned. However, I believe I can explain how to arrive at the correct answer in a much easier to follow way. The trick lies in the value of the variable k. Recall that when you're looking for an nth root, k has values from 0 to n-1. In this case, you're after a fifth root, so k ranges from 0 to 4. You have a value for theta that depends on k, so let's plug in the values for k and see what happens:

θ=54+k72\displaystyle \theta =54^\circ +k\cdot 72^\circ

θ=54+072=54\displaystyle \theta =54^\circ +0\cdot 72^\circ =54^\circ (k = 0)
θ=54+172=126\displaystyle \theta =54^\circ +1\cdot 72^\circ =126^\circ (k = 1)
θ=54+272=198\displaystyle \theta =54^\circ +2\cdot 72^\circ =198^\circ (k = 2)
θ=54+372=270\displaystyle \theta =54^\circ +3\cdot 72^\circ =270^\circ (k = 3)
θ=54+472=342\displaystyle \theta =54^\circ +4\cdot 72^\circ =342^\circ (k = 4)

Now, the instructions say to find a theta strictly between 270 and 360 degrees, or: 270<θ<360\displaystyle 270^\circ <\theta <360^\circ. When k = 3, theta is exactly 270. But you have a strictly less than sign, not a less than or equal to sign, so exactly 270 degrees is not allowed. But the next value up, when k = 4, is valid. So, that's your solution.
 
I have to be honest, I haven't any earthly idea what your web program is trying to explain there. It might as well be an alien language, as far as I'm concerned. However, I believe I can explain how to arrive at the correct answer in a much easier to follow way. The trick lies in the value of the variable k. Recall that when you're looking for an nth root, k has values from 0 to n-1. In this case, you're after a fifth root, so k ranges from 0 to 4. You have a value for theta that depends on k, so let's plug in the values for k and see what happens:

θ=54+k72\displaystyle \theta =54^\circ +k\cdot 72^\circ

θ=54+072=54\displaystyle \theta =54^\circ +0\cdot 72^\circ =54^\circ (k = 0)
θ=54+172=126\displaystyle \theta =54^\circ +1\cdot 72^\circ =126^\circ (k = 1)
θ=54+272=198\displaystyle \theta =54^\circ +2\cdot 72^\circ =198^\circ (k = 2)
θ=54+372=270\displaystyle \theta =54^\circ +3\cdot 72^\circ =270^\circ (k = 3)
θ=54+472=342\displaystyle \theta =54^\circ +4\cdot 72^\circ =342^\circ (k = 4)

Now, the instructions say to find a theta strictly between 270 and 360 degrees, or: 270<θ<360\displaystyle 270^\circ <\theta <360^\circ. When k = 3, theta is exactly 270. But you have a strictly less than sign, not a less than or equal to sign, so exactly 270 degrees is not allowed. But the next value up, when k = 4, is valid. So, that's your solution.

Can I kiss you???
Thankssssssssssssssssssss
 
[FONT=KaTeX_Main][FONT=KaTeX_Math]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]
Do not know why you make this so complicated?

z5=7776cis(π2)=7776cos(π2)+7776isin(π2)\displaystyle {z^5} = 7776cis\left( {\dfrac{{ - \pi }}{2}} \right) =7776 \cos \left( {\dfrac{{ - \pi }}{2}} \right) +7776 i\sin \left( {\dfrac{{ - \pi }}{2}} \right)

So z=77765cis(π10)\displaystyle z = \sqrt[5]{{7776}}cis\left( {\dfrac{{ - \pi }}{{10}}} \right). Because π10IV\displaystyle \dfrac{{ - \pi }}{{10}} \in IV you are done!
 
Do not know why you make this so complicated?

z5=7776cis(π2)=7776cos(π2)+7776isin(π2)\displaystyle {z^5} = 7776cis\left( {\dfrac{{ - \pi }}{2}} \right) =7776 \cos \left( {\dfrac{{ - \pi }}{2}} \right) +7776 i\sin \left( {\dfrac{{ - \pi }}{2}} \right)

So z=77765cis(π10)\displaystyle z = \sqrt[5]{{7776}}cis\left( {\dfrac{{ - \pi }}{{10}}} \right). Because π10IV\displaystyle \dfrac{{ - \pi }}{{10}} \in IV you are done!

Sorry but i think i don't get it.
So what is the real part and imaginary part from your solution.
 
#update 1: Thanks ksdhart for your solution. I got what you mean. But I got another problem. The way the explain so hard to understand (khanacademy)
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(3*O)] = 27 [cos (k*360) + isin(k*360)] ===> 3*O = k*360
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
Find the solution of the following equation whose argument is strictly between 270, degree and ​​360, degree.
Z = A + Bi

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 270 decrees and 360 decrees.Therefore, we need to find the multiple of 90 decrees that is strictly within the range of 270 decrees -54 decrees =216 decrees & 360-54=306 (decrees). 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)
The above contains bits and pieces from at least four different conversations. Due to your editing, instead of replying, it now makes zero sense, and the thread is (for me, at least) completely lost.

Please reply (at the bottom) with your current question, showing your work so far. Thank you.
 
Sorry but i think i don't get it.
So what is the real part and imaginary part from your solution.

Pka's solution is written in polar form, rather than the "standard" notation that you're used to seeing. You should recall from your lessons how to write a complex number in polar form. Suppose you have a complex number z = a + bi. Then you could write it in polar form as:

z=r[cos(θ)+isin(θ)]\displaystyle z=r\left[cos\left(\theta \right)+i\cdot sin\left(\theta \right)\right]

Where:

r=a2+b2\displaystyle r=\sqrt{a^2+b^2} and θ=tan1(ba)\displaystyle \theta =tan^{-1}\left(\frac{b}{a}\right)

More specifically, you'd want to use a variant of the arctangent. In computer programming circles, this variant is known as atan2(x,y). See this Wikipedia page for details about how it works.

Based on the above, can you see what the values of r and theta might be in pka's solution? And from that, can you work backwards to determine a and b?

z=77765[cos(18)+isin(18)]\displaystyle z=\sqrt[5]{7776}\cdot \left[cos\left(18^\circ \right)+i\cdot sin\left(18^\circ \right)\right]
 
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