Help With Complex Numbers #2

[jisbon]

Homework Statement

NIL

Relevant Equations

NIL

--Continued--

Spoiler: Solved

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

 

[PeroK]

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.

 

[jisbon]

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]

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]

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.

 

[jisbon]

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]

$Ax+B= -(3\overline{z}+3z)x+z\overline{z}$
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:

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

$x^2 + \frac{A}{3}x + \frac{B}{3} = (x - z)(x - \overline{z})$

 

[jisbon]

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

 

[jisbon]

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 :/

 

[jisbon]

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]

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]

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 $\theta$.

 

[jisbon]

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]

Your solution to 6 is correct.

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

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

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

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

 

[jisbon]

Your solution to 6 is correct.

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

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

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

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

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 :/

 

[jisbon]

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]

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?

 

[jisbon]

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]

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

No, but it doesn't matter. Instead, think geometric series. That's the key.

 

[jisbon]

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

 

[PeroK]

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

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.

 

[jisbon]

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]

I'm leaving out the k=0 where $x^n =1$ to make things easier later, so

That's making it harder!

 

[WWGD]

Think of properties of all n-th roots of unity.

 

[jisbon]

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.

 

[jisbon]

Been thinking another approach. Since
$\sum_{k=0}^9 x^k= 0$
$z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 =0$
$re^{9i\theta} + re^{8i\theta} + re^{7i\theta} + re^{6i\theta} + re^{5i\theta} + re^{4i\theta} + re^{3i\theta} + re^{2i\theta} + re^{i\theta} + 1 = 0$
EDIT: Thought further on this, does this mean:

$cos9\theta + cos8\theta + cos7\theta + cos6\theta + cos5\theta + cos4\theta + cos3\theta + cos2\theta + cos\theta + 1 = 0$
and
$sin9\theta + sin8\theta + sin7\theta + sin6\theta + sin5\theta + sin4\theta + sin3\theta + sin2\theta + sin\theta = 0$
?

 

[FactChecker]

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.

I could not find anything that said that A and B were real. I think we are assuming it because otherwise there is no unique solution and the problem would not be a very good one. (Or it is stated somewhere earlier that I could not find. In these multiple threads, I could never find the full original statement of the problem.)

 

[PeroK]

Only property I could think of is that they will all sum to zero eventually :/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.

First, you could always look up the sum of a geometric series. Second, you really should be able to do it for yourself. Third, @HallsofIvy already did it for you.

In any case, you have:

$1 + x + \dots + x^9 = \frac{1-x^{10}}{1-x}$

That's the key.