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Help With Complex Numbers #2

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Homework Statement
NIL
Homework Equations
NIL
--Continued--
4)
Take ##3+7i## is a solution of ##3x^2+Ax+B=0##
Since ##3+7i## is a solution, I can only gather :
##(z−(3+7i))(...)=3x2+Ax+B##
Not sure on how to go from here.
EDIT: I got A =18 and B=174, is this correct?
I recognized that since there's a 3, this means the other root must be a conjugate, hence
##(z-(3+7i))(z-(3-7i))##
##(z-3)^2-(7i)^2 =0##
##z^2+6z+58=0##
##3z^2+18z+174=0##

6)
Suppose ##z=2e^{ikπ}##and
##z^{n}=2^5 e^{iπ/8}##
Find k such that z has smallest positive argument?

I don't understand this question :/ For z to have smallest positive principal argument, what does it entail/mean?
EDIT: Tried again. Got the following:
##z^{n}=2^n e^{inkπ} = 2^5 e^{iπ/8}##
## nk = 1/8##
##5k =1/8##
##k = 1/40##?

7)
Let
##\sum_{k=0}^9 x^k = 0##
Find smallest positive argument. Same thing as previous question, but I guess I can expand to
##z+z_{2}+z_{3}+...+z_{9}=0##
##z=re^{iθ}##
##re^{iθ}+re^{2iθ}+re^{3iθ}+...##
What do I do to proceed on?
Cheers
 
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PeroK

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Homework Statement: NIL
Homework Equations: NIL

--Continued--

4)
Take ##3+7i## is a solution of ##3x^2+Ax+B=0##
Since ##3+7i## is a solution, I can only gather :
##(z−(3+7i))(...)=3x2+Ax+B##
Not sure on how to go from here.
EDIT: I got A =18 and B=174, is this correct?
I recognized that since there's a 3, this means the other root must be a conjugate, hence
##(z-(3+7i))(z-(3-7i))##
##(z-3)^2-(7i)^2 =0##
##z^2+6z+58=0##
##3z^2+18z+174=0##
I would take a more general approach to this. You are given that:

##3x^2 + Ax + B = 3(x - z)(x - \overline{z})##

Where ##z = 3 + 7i##.

Try expressing ##A, B## in terms of ##z## in general first. And see what you get.
 
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I would take a more general approach to this. You are given that:

3x2+Ax+B=3(x−z)(x−¯¯¯z)3x2+Ax+B=3(x−z)(x−z¯)

Where z=3+7iz=3+7i.

Try expressing A,BA,B in terms of zz in general first. And see what you get.
I got:

3x2+Ax+B=3(x2−x¯¯¯z−xz+z¯¯¯z)3x2+Ax+B=3(x2−xz¯−xz+zz¯)
Ax+B=−(3¯¯¯z+z)x+z¯¯¯zAx+B=−(3z¯+z)x+zz¯
SoA=−(3¯¯¯z+z)A=−(3z¯+z) and B=z¯¯¯zB=zz¯
 

PeroK

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I got:

##3x^2+Ax+B = 3 (x^2 -x\overline{z}-xz+z\overline{z})##
##Ax+B= -(3\overline{z}+z)x+z\overline{z}##
So## A =-(3\overline{z}+z)## and ##B= z\overline{z}##
That's not quite right. You need to be more careful.
 

HallsofIvy

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I am puzzled by your statement "since there's a 3, this means the other root must be a conjugate". The other root must be the complex conjugate of 3+ 7i, 3- 7i, because all coefficients are real. Because the leading coefficient is 3, the polynomial must be 3(x-(3+7i))(x- (3- 7i))= 3((x-3)- 7i)((x-3)+ 7i)= 3((x- 3)^2+ 49)= 3(x^2- 6x+ 58)= 3x^2- 18x+ 174.
 
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That's not quite right. You need to be more careful.
##Ax+B= -(3\overline{z}+3z)x+z\overline{z}##
So from here, I can easily get B since it's ##a^2+b^2##, which gives me 54.
 

PeroK

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##Ax+B= -(3\overline{z}+3z)x+z\overline{z}##
So from here, I can easily get B since it's ##a^2+b^2##, which gives me 54.
You do need to be a lot more careful. I would get rid of the ##3## first:

##3x^2 + Ax + B = 3(x - z)(x - \overline{z})##

##x^2 + \frac{A}{3}x + \frac{B}{3} = (x - z)(x - \overline{z})##
 
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You do need to be a lot more careful. I would get rid of the ##3## first:

##3x^2 + Ax + B = 3(x - z)(x - \overline{z})##

##x^2 + \frac{A}{3}x + \frac{B}{3} = (x - z)(x - \overline{z})##
Solving it, it seems my original answer was wrong. From your equation, I got A = -18 instead of 18. B is still 174 though
 
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I am puzzled by your statement "since there's a 3, this means the other root must be a conjugate". The other root must be the complex conjugate of 3+ 7i, 3- 7i, because all coefficients are real. Because the leading coefficient is 3, the polynomial must be 3(x-(3+7i))(x- (3- 7i))= 3((x-3)- 7i)((x-3)+ 7i)= 3((x- 3)^2+ 49)= 3(x^2- 6x+ 58)= 3x^2- 18x+ 174.
Yea, I think I phrased myself wrongly haha :/
 
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You do need to be a lot more careful. I would get rid of the ##3## first:

##3x^2 + Ax + B = 3(x - z)(x - \overline{z})##

##x^2 + \frac{A}{3}x + \frac{B}{3} = (x - z)(x - \overline{z})##
Thanks for the help :) Mind checking 6? I understand that 6 and 7 have some similarities, but I can't seem to get 7 (if my 6 is even right)
 

PeroK

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Solving it, it seems my original answer was wrong. From your equation, I got A = -18 instead of 18. B is still 174 though
The point is that there are advantages in getting a general expression:

##\frac{A}{3} = b = -(z + \overline{z}) = -2Re(z)## and ##\frac{B}{3} = c = z \overline{z} = |z|^2##

This allows you to read off the answers for whatever ##z## you are given. You let the algebra do the work, rather than fighting with specific numbers. In this case, ##z= 3 + 7i## was quite simple. But, if you'd been given ##z = 3.7 + 7.5i## or somthing even worse, then the benefits of deriving the expression generally become very significant.

Also, as you get more experienced, it's things like ##z + \overline{z} = 2 Re(z)## that ought to stick in your mind. That's when you become more fluent and confident. Hammering away with numbers all the time leads to little if any pattern recognition.
 

PeroK

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6)
Suppose ##z=2e^{ikπ}##and
##z^{n}=2^5 e^{iπ/8}##
Find k such that z has smallest positive argument
I don't understand this question :/ For z to have smallest positive principal argument, what does it entail/mean?
EDIT: Tried again. Got the following:
##z^{n}=2^n e^{inkπ} = 2^5 e^{iπ/8}##
## nk = 1/8##
##5k =1/8##
##k = 1/40##?
I don't understand what this question is asking.

7)
Let
##\sum_{k=0}^9 x^k = 0##
Find smallest positive argument. Same thing as previous question, but I guess I can expand to
##z+z_{2}+z_{3}+...+z_{9}=0##
##z=re^{iθ}##
##re^{iθ}+re^{2iθ}+re^{3iθ}+...##
What do I do to proceed on?
Cheers
I guess here ##x## is a complex number. In any case, you have a 9th degree polynomial that will have 9 complex roots. You need to find the one with the smallest argument. I.e. smallest ##\theta## in the polar form.

Or, to be precise, the smallest non-zero ##\theta##.
 
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I don't understand what this question is asking.



I guess here ##x## is a complex number. In any case, you have a 9th degree polynomial that will have 9 complex roots. You need to find the one with the smallest argument. I.e. smallest ##\theta## in the polar form.

Or, to be precise, the smallest non-zero ##\theta##.
For Q6, I'm confused by this too. Anyone else can shine a light on this? Haha

Regarding the last question,yes x is a complex number. how do I exactly find the smallest angle? Do I solve for x first in the equation?
 

HallsofIvy

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Your solution to 6 is correct.

For 7 you have [tex]\sum_{k=0}^9 x^k= 0[/tex] and then [tex]z= re^{i\theta}[/tex] (surely, your "x" and "z" should be the same!) and then [tex]re^{i\theta}+ r^2e^{i\theta}+ r^3e^{i\theta}+ \cdot\cdot\cdot[/tex].

There are several things wrong with that! First, it starts with k= 1 rather than k= 0. When k= 0, [tex]r^0e^{i(0)\theta}= 1[/tex]. Also, [tex](re^{i\theta})^n= r^n e^{in\theta}[/tex]. You forgot the "n" exponent on r. Finally, the "[tex]\cdot\cdot\cdot[/tex]" on the end implies the sum continues to infinity. This sum only goes to k= 9. You should have [tex]1+ re^{i\theta}+ r^2e^{2i\theta}+ r^3e^{3i\theta}+ \cdot\cdot\cdot+ r^9e^{9i\theta}[/tex].

However, the key point is that [tex]\sum_{i=0}^\infty x^k= 1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9[/tex] is a geometric sum. Write [tex]S= 1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9[/tex]. Then [tex]S- 1= x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9= x(1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8)[/tex]

Add and subtract [tex]x^9[/tex] inside the parenthese:
[tex]S- 1= x(1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9- x^9)= x(1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9)- x^{10}[/tex]. We can write that as [tex]S-1= xS- x^{10}[/tex] so that [tex]S- xS= S(1- x)= 1- x^{10}[/tex] so [tex]S= \frac{1- x^{10}}{1- x}[/tex]. With [tex]x= re^{i\theta}[/tex], [tex]x^{10}= r^{10}e^{10i\theta}[/tex] so [tex]S= \frac{1- r^{10}e^{10i\theta}}{1- re^{i\theta}[/tex].
 
257
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Your solution to 6 is correct.

For 7 you have [tex]\sum_{k=0}^9 x^k= 0[/tex] and then [tex]z= re^{i\theta}[/tex] (surely, your "x" and "z" should be the same!) and then [tex]re^{i\theta}+ r^2e^{i\theta}+ r^3e^{i\theta}+ \cdot\cdot\cdot[/tex].

There are several things wrong with that! First, it starts with k= 1 rather than k= 0. When k= 0, [tex]r^0e^{i(0)\theta}= 1[/tex]. Also, [tex](re^{i\theta})^n= r^n e^{in\theta}[/tex]. You forgot the "n" exponent on r. Finally, the "[tex]\cdot\cdot\cdot[/tex]" on the end implies the sum continues to infinity. This sum only goes to k= 9. You should have [tex]1+ re^{i\theta}+ r^2e^{2i\theta}+ r^3e^{3i\theta}+ \cdot\cdot\cdot+ r^9e^{9i\theta}[/tex].

However, the key point is that [tex]\sum_{i=0}^\infty x^k= 1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9[/tex] is a geometric sum. Write [tex]S= 1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9[/tex]. Then [tex]S- 1= x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9= x(1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8)[/tex]

Add and subtract [tex]x^9[/tex] inside the parenthese:
[tex]S- 1= x(1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9- x^9)= x(1+ x+ x^2+ x^3+ x^4+ x^5+ x^6+ x^7+ x^8+ x^9)- x^{10}[/tex]. We can write that as [tex]S-1= xS- x^{10}[/tex] so that [tex]S- xS= S(1- x)= 1- x^{10}[/tex] so [tex]S= \frac{1- x^{10}}{1- x}[/tex]. With [tex]x= re^{i\theta}[/tex], [tex]x^{10}= r^{10}e^{10i\theta}[/tex] so [tex]S= \frac{1- r^{10}e^{10i\theta}}{1- re^{i\theta}[/tex].
Hi there.
Thanks for taking your time to type this chunk out :0
Unfortunately, I don't really understand how the summation (##S= \frac{1- r^{10}e^{10i\theta}}{1- re^{i\theta}}##) can lead to finding the smallest argument :/
 
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Hold on.
With the summation
(##S= \frac{1- r^{10}e^{10i\theta}}{1- re^{i\theta}}##)
Since it equates to 0, I can assume
##1- r^{10}e^{10i\theta}## = 0
##r^{10}e^{10i\theta}## = 1
##re^{i\theta}## = 1
Ok I'm stuck :/
 

PeroK

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For Q6, I'm confused by this too. Anyone else can shine a light on this? Haha

Regarding the last question,yes x is a complex number. how do I exactly find the smallest angle? Do I solve for x first in the equation?
This looks considerably harder. My first thought is to look at the problem for lower orders of the polynomial. What happens for 3, 5, 7 and then finally 9? Is there some sort of pattern?
 
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This looks considerably harder. My first thought is to look at the problem for lower orders of the polynomial. What happens for 3, 5, 7 and then finally 9? Is there some sort of pattern?
3,5,7,9
Do you mean by ##r^2 e^{i2\theta}## , ##r^4 e^{i4\theta}## etc..? Not sure what you meant by pattern too
 

PeroK

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Do you mean by ##r^2 e^{i2\theta}## , ##r^4 e^{i4\theta}## etc..? Not sure what you meant by pattern too
No, but it doesn't matter. Instead, think geometric series. That's the key.
 
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No, but it doesn't matter. Instead, think geometric series. That's the key.
I did learn that a summation of a GP is ##S= \frac{a}{1-r}##
The problem is I don't understand how does GP helps in solving this problem here o_O
 

PeroK

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I did learn that a summation of a GP is ##S= \frac{a}{1-r}##
The problem is I don't understand how does GP helps in solving this problem here o_O
We can't do your homework for you. You have to be able to follow a line or argument for yourself. At the moment you demonstrate an ability to think only 1-2 steps. You need to get used to thinking further. It doesn't matter if it doesn't lead anywhere, you just rewind and try again.

You should spend maybe 15-30 mins on this, trying everything you can think of.
 
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So far what I understood is:
##\sum_{k=0}^9 x^k= 0##
So yes I do know this is a GP equation where:
##S= \frac{a(1-r^n)}{1-r}##
I'm leaving out the k=0 where ##x^n =1## to make things easier later, so
Where ##a = x## , ##r = k## , ##n=9##
##S= 1+ \frac{x(1-k^9)}{1-k}##
..
 

PeroK

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WWGD

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Think of properties of all n-th roots of unity.
 
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Think of properties of all n-th roots of unity.
Only property I could think of is that they will all sum to zero eventually :/

That's making it harder!
If I used the original stuff, I will get:
##\frac{z^k (1-k)^9}{1-k}## = ##\frac{re^{ki\theta} (1-k)^9}{1-k}## = ##re^{ki\theta} (1-k)^8## = ##(1-k)^8## since k=0 for the first value. Think I made a mistake because it seems to far off from what @HallsofIvy provided

EDIT: Yep definitely made a mistake trying to figure out where.
 
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