Help with Complex Numbers

[jisbon]

Homework Statement

NIL

Relevant Equations

NIL

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 :)

Spoiler: Solved Questions

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∣

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

 

[PeroK]

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

What's the relationship between $a + bi$ and $a^2 + b^2$?

 

[jisbon]

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 :)
$a^2 + b^2$ = $(a + bi)(a-bi)$

 

[PeroK]

Okay :)
$a^2 + b^2$ = $(a + bi)(a-bi)$

And what does that represent?

 

[jisbon]

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]

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?

 

[jisbon]

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]

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?

 

[jisbon]

Actually, that's the modulus squared.

What properties does the modulus have?

Is the answer $80$? :)
EDIT: 80 I mean

 

[PeroK]

Is the answer $80$? :)
EDIT: 80 I mean

Yes.

 

[jisbon]

Yes.

Thanks :) Any clues for the next question :>

 

[PeroK]

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?

 

[jisbon]

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]

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.

 

[jisbon]

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]

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.

 

[jisbon]

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]

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.

 

[jisbon]

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]

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]

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.

 

[jisbon]

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]

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?

 

[jisbon]

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]

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.

 

[Delta2]

for 5, instead of doing it the hard way (by replacing $z_1=a+bi, z_2=c+di$ e.t.c) use properties of the modulus and algebra in the set of complex numbers. Properties of modulus are pretty much like the properties of absolute value in real numbers and algebra in C is pretty much the same as algebra in R so for example
$\frac{1}{z_1}+\frac{1}{z_2}=\frac{z_1+z_2}{z_1z_2}$

 

[jisbon]

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.

Assuming b =0, it does

 

[jisbon]

for 5, instead of doing it the hard way (by replacing $z_1=a+bi, z_2=c+di$ e.t.c) use properties of the modulus and algebra in the set of complex numbers. Properties of modulus are pretty much like the properties of absolute value in real numbers and algebra in C is pretty much the same as algebra in R so for example
$\frac{1}{z_1}+\frac{1}{z_2}=\frac{z_1+z_2}{z_1z_2}$

So for question 5, answer will be 2/49?
Thanks for the tip :)

 

[PeroK]

Assuming b =0, it does

You don't have to assume $b = 0$. You should be able to show that $b = 0$ and $a \ge 0$.

 

[Delta2]

So for question 5, answer will be 2/49?
Thanks for the tip :)

Yes that's correct.

 

[jisbon]

You don't have to assume $b = 0$. You should be able to show that $b = 0$ and $a \ge 0$.

Showing b=0 is done on the very first part where there is no imaginary number on the right is it?

 

[PeroK]

Showing b=0 is done on the very first part where there is no imaginary number on the right is it?

Yes.

 

[jisbon]

Yes.

So since I managed to show, I can just sub in:
a+bi+7(a−bi)=∣(a+4)+bi∣
where b =0,
4/7 + 7(4/7) = (4/7) +4
which is true.
Hence z = 4/7 + 0i?

 

[PeroK]

So since I managed to show, I can just sub in:
a+bi+7(a−bi)=∣(a+4)+bi∣
where b =0,
4/7 + 7(4/7) = (4/7) +4
which is true.
Hence z = 4/7 + 0i?

What I would have done is noted that $|z + 4| = c$, where $c \ge 0$.

That shows that $b = 0$ and $a \ge 0$ (do the algebra). You don't have to guess or assume that $b = 0$.

Also, a little trick, as $a \ge 0$, we have $|z + 4| = |a + 4| = a + 4$

And that avoids having to square the equation.

 

[jisbon]

What I would have done is noted that $|z + 4| = c$, where $c \ge 0$.

That shows that $b = 0$ and $a \ge 0$. You don't have to guess or assume that $b = 0$.

Also, a little trick, as $a \ge 0$, we have $|z + 4| = |a + 4| = a + 4$

And that avoids having to square the equation.

Oh okay. So by proving b=0, we can now assume z = 4/7 +0i right?
Also for the next question, I've edited to include my initial answers. Do you think they are correct? Thanks.