Help With Complex Numbers #2

jisbon

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

Last edited:
Related Precalculus Mathematics Homework Help News on Phys.org

PeroK

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member
2018 Award
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

Homework Helper
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

Homework Helper
Gold Member
2018 Award
$Ax+B= -(3\overline{z}+3z)x+z\overline{z}$
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

• PeroK

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

Homework Helper
Gold Member
2018 Award
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.

• Emmo Amaranth and jisbon

PeroK

Homework Helper
Gold Member
2018 Award
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

Homework Helper
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

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

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member
2018 Award
I'm leaving out the k=0 where $x^n =1$ to make things easier later, so
That's making it harder!

WWGD

Gold Member
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.

Last edited:

"Help With Complex Numbers #2"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving