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Help with Complex Numbers

  • Thread starter jisbon
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Hello all!
Thanks for helping me out so far :) Really appreciate it.
I don't seem to understand some of the questions presented to me, so if anyone has an idea on how to start the questions, please do render your assistance :)
1)
Suppose
(a+bi)(c+di)(e+fi)=4+8i(a+bi)(c+di)(e+fi)=4+8i
Find the value of
(a2+b2)(c2+d2)(e2+f2)(a2+b2)(c2+d2)(e2+f2)
Not sure what I'm suppose to do here, expanding is probably out of the question, does squaring (a+bi)(c+di)(e+fi)(a+bi)(c+di)(e+fi) helps to find out (a2+b2)(c2+d2)(e2+f2)(a2+b2)(c2+d2)(e2+f2)?

2)
Let
S=(cos(π/5)+isin(π/5))n,nϵNS=(cos(π/5)+isin(π/5))n,nϵN
What I understand here is that I'm supposed to find the amount of distinct roots in this equation? How do I even start?

5)
Let ∣z1∣=∣z2∣=7∣z1∣=∣z2∣=7
If ∣z1+z2∣=2∣z1+z2∣=2,solve ∣1/z1+1/z2∣
1567165455919.png


How do I even proceed from here?

3)
Let z be complex number that allows:
z+7¯¯¯z=∣¯¯¯z+4∣∣z+7z¯=∣z¯+4∣
Find z.
My working:
a+bi+7(a−bi)=∣(a+4)+bi∣a+bi+7(a−bi)=∣(a+4)+bi∣
8a−6bi=√(a+4)2+b28a−6bi=√(a+4)2+b2
64a2−96abi−36b2=a2+8a+16+b264a2−96abi−36b2=a2+8a+16+b2
Not sure where to proceed from here.
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+z2+z3+...+z9=0z+z2+z3+...+z9=0
##z=re^iθ##
##rei^θ+re^2iθ+re^3iθ+...##
What do I do to proceed on?
Cheers
 
Last edited:

PeroK

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You need to take these one at a time. Let's start with 1):

What's the relationship between ##a + bi## and ##a^2 + b^2##?
 
257
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You need to take these one at a time. Let's start with 1):

What's the relationship between ##a + bi## and ##a^2 + b^2##?
Okay :)
##a^2 + b^2## = ##(a + bi)(a-bi)##
 
257
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And what does that represent?
Not sure what you mean there. But I do get that the solution will be
##(4+8i)(a-bi)(c-di)(e-fi)##?
 

PeroK

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Not sure what you mean there. But I do get that the solution will be
##(4+8i)(a-bi)(c-di)(e-fi)##?
Have you ever heard of the "modulus" of a complex number?
 
257
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Have you ever heard of the "modulus" of a complex number?
Oh yes, "modulus" of a complex number will be ##a^2+b^2##
But how does it help me ?
 

PeroK

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Oh yes, "modulus" of a complex number will be ##a^2+b^2##
But how does it help me ?
Actually, that's the modulus squared.

What properties does the modulus have?
 
257
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Actually, that's the modulus squared.

What properties does the modulus have?
Is the answer ## 80##? :)
EDIT: 80 I mean
 

PeroK

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2)
Let
##S=\{(cos(π/5)+isin(π/5))^n \ ,nϵN \}##
What I understand here is that I'm supposed to find the amount of distinct roots in this equation? How do I even start?
For number 2), your original statement was better. I've tidied it up. ##S## is a set. On the face of it, ##S## has an infinite numbers of elements. But, are some of the elements the same? If so, how many distinct elements does it have?
 
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For number 2), your original statement was better. I've tidied it up. ##S## is a set. On the face of it, ##S## has an infinite numbers of elements. But, are some of the elements the same? If so, how many distinct elements does it have?
So from what you state, I should probably start listing down all the possible elements?
Since ##\theta## can only go from ##-\pi## to ##\pi##
If I make it to :
##re^{i\frac{\pi}{5}}##
Then there will be:
##re^{i\frac{2\pi}{5}}##
##re^{i\frac{3\pi}{5}}##
##re^{i\frac{4\pi}{5}}##
##re^{i\frac{5\pi}{5}}##
and the negative.
So total 10?
 

PeroK

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So from what you state, I should probably start listing down all the possible elements?
Since ##\theta## can only go from ##-\pi## to ##\pi##
If I make it to :
##re^{i\frac{\pi}{5}}##
Then there will be:
##re^{i\frac{2\pi}{5}}##
##re^{i\frac{3\pi}{5}}##
##re^{i\frac{4\pi}{5}}##
##re^{i\frac{5\pi}{5}}##
and the negative.
So total 10?
You could do a better job of explaining that, but ##10## is correct.
 
257
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You could do a better job of explaining that, but ##10## is correct.
Oopsie haha, ok at least I kind of understood what it means.
How about 3? I'm kind of confused because I only have 1 equation when I need to solve for both a and b :(
 

PeroK

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Oopsie haha, ok at least I kind of understood what it means.
How about 3? I'm kind of confused because I only have 1 equation when I need to solve for both a and b :(
For number 3, does anything look strange about that equation? You need to think out of the box a little.
 
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For number 3, does anything look strange about that equation? You need to think out of the box a little.
There seems to be no imaginary number on the right hand side, does it necessary mean ##-96ab = 0##?
 

PeroK

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There seems to be no imaginary number on the right hand side, does it necessary mean ##-96ab = 0##?
Yes, the RHS is a non-negative real number. What does that say about ##z##? Hint: start again.
 
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Yes, the RHS is a non-negative real number. What does that say about ##z##? Hint: start again.
This means that z is solely a real number, is it?
Also not sure what you meant by 'starting' again. Is it rewriting the equations without ##ib## involved?
 

PeroK

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This means that z is solely a real number, is it?
Also not sure what you meant by 'starting' again. Is it rewriting the equations without ##ib## involved?
Yes, you either see directly that ##z## is real, or set ##z = a + ib## and show that ##b = 0##.
 

PeroK

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Yes, you either see directly that ##z## is real, or set ##z = a + ib## and show that ##b = 0##.
PS what you did wasn't wrong, but it got too complicated, which was a clue that you'd missed something. In these cases, it's usually better to start again than to continue with your complicated equations.
 
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Yes, you either see directly that ##z## is real, or set ##z = a + ib## and show that ##b = 0##.
Restarting the equation without ##ib## involved gives me:
##a+7a = a^2+8a+16##
##a^2 +16 =0##
##a = -4i##?
Now it seems to be a problem :/

PS: Nevermind, I think I got a mistake. Is ##a = 4/7## ? Hence ## z= 4/7 + 0i##
 

PeroK

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Restarting the equation without ##ib## involved gives me:
##a+7a = a^2+8a+16##
##a^2 +16 =0##
##a = -4i##?
Now it seems to be a problem :/

PS: Nevermind, I think I got a mistake. Is ##a = 4/7## ? Hence ## z= 4/7 + 0i##
Did you try that in the original equation?
 
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Did you try that in the original equation?
I think I messed up my equation.
Is it supposed to be:
##a+7a = a+4## now? Since I square and square root the RHS, giving me back ##a+4##?
 

PeroK

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I think I messed up my equation.
Is it supposed to be:
##a+7a = a+4## now? Since I square and square root the RHS, giving me back ##a+4##?
You don't have to square the equation. In any case, you should check whether ##z = 4/7## is a solution by direct substitution into the original equation.
 

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