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complex numbers problems

i have a couple quesitons im stuck on for complex numbers.
https://imgur.com/a/CLUuecy

for the first one, stuck on second question, (w^2) im unsure what its actually asking, is it asking for the distance from the point of intersection with the arg and the circle, to the origin or to (3,0)?

for the second one ive tried a few things and got that there is another equation z^2 +4z + p + 12 but i cant figure out what to do after that, i try using roots of polynomials but i just end up in a dead end.
|w|^2 is distance (squared) between w and the origin.

For the other question, I would start from the observation that since your poly is real, the roots must come in conjugate pairs. So you can write z2 = a+ib and z3 = a - ib (for real numbers a, b that you have to find).

You can also find z1+z2+z3 from the coefficient of x^2, which will determine a.

The area of the triangle will then give you b, and then knowing the 3 roots gives you p and q.
Original post by DFranklin
|w|^2 is distance (squared) between w and the origin.

For the other question, I would start from the observation that since your poly is real, the roots must come in conjugate pairs. So you can write z2 = a+ib and z3 = a - ib (for real numbers a, b that you have to find).

You can also find z1+z2+z3 from the coefficient of x^2, which will determine a.

The area of the triangle will then give you b, and then knowing the 3 roots gives you p and q.

ok, ive got it so 3a^2 + 3b^2 = 35 after doing (a+ib)(a-ib)(3) = 35, what do i do next?
Original post by Gent2324
ok, ive got it so 3a^2 + 3b^2 = 35 after doing (a+ib)(a-ib)(3) = 35, what do i do next?


Original post by DFranklin
You can also find z1+z2+z3 from the coefficient of x^2, which will determine a.

If you know a, and you know 3a^2+3b^2 = 35, then you know b... (well, you know b^2, and the sign of b doesn't matter).
Original post by DFranklin
|w|^2 is distance (squared) between w and the origin.

For the other question, I would start from the observation that since your poly is real, the roots must come in conjugate pairs. So you can write z2 = a+ib and z3 = a - ib (for real numbers a, b that you have to find).

You can also find z1+z2+z3 from the coefficient of x^2, which will determine a.

The area of the triangle will then give you b, and then knowing the 3 roots gives you p and q.



Original post by DFranklin
If you know a, and you know 3a^2+3b^2 = 35, then you know b... (well, you know b^2, and the sign of b doesn't matter).

wait i got it wrong, i multiplied them all together = area which obviously is wrong, so how do we actually know 3a^2 + 3b^2 = 35? and i wasnt letting z1 = 3, a was just a part of the second root a+ib
Original post by Gent2324
wait i got it wrong, i multiplied them all together = area which obviously is wrong, so how do we actually know 3a^2 + 3b^2 = 35? and i wasnt letting z1 = 3, a was just a part of the second root a+ib


One of the roots is z1=3z_1 = 3. Then since all coeffs of the polynomial are real, that means z2,z3z_2, z_3 are complex conjugates of each other. Geometrically, this means z2z_2 is a reflection of z3z_3 in the real line. A quick sketch would show you that the triangle formed is actually an isosceles triangle whose area is given by (3−Rez2)×Imz2(3-\mathrm{Re} z_2) \times \mathrm{Im} z_2 where 3−Rez3-\mathrm{Re} z is the 'height' of the triangle form the base to the opposite vertex z1=3z_1 = 3. And then Imz2\mathrm{Im} z_2 is the length from one of the triangle's other vertices, to the line of symmetry. (Diagram below)

If you choose to denote z2=a+ibz_2=a+ib as DFranklin has, then Rez2=a\mathrm{Re} z_2 = a and Imz2=b\mathrm{Im} z_2 = b hence you get the eq.

(3−a)b=35(3-a)b = 35

from which you can answer the question.



Original post by RDKGames
One of the roots is z1=3z_1 = 3. Then since all coeffs of the polynomial are real, that means z2,z3z_2, z_3 are complex conjugates of each other. Geometrically, this means z2z_2 is a reflection of z3z_3 in the real line. A quick sketch would show you that the triangle formed is actually an isosceles triangle whose area is given by (3−Rez2)×Imz2(3-\mathrm{Re} z_2) \times \mathrm{Im} z_2 where 3−Rez3-\mathrm{Re} z is the 'height' of the triangle form the base to the opposite vertex z1=3z_1 = 3. And then Imz2\mathrm{Im} z_2 is the length from one of the triangle's other vertices, to the line of symmetry. (Diagram below)

If you choose to denote z2=a+ibz_2=a+ib as DFranklin has, then Rez2=a\mathrm{Re} z_2 = a and Imz2=b\mathrm{Im} z_2 = b hence you get the eq.

(3−a)b=35(3-a)b = 35

from which you can answer the question.




ah right, i get you all up to the equation with a and b, but i still dont get how i can solve it? theres 2 unknowns and only 1 equation?
Original post by Gent2324
ah right, i get you all up to the equation with a and b, but i still dont get how i can solve it? theres 2 unknowns and only 1 equation?


But you should be able to determine what aa is from considering the sum of the roots, which is what DFranklin has been suggesting.
Original post by Gent2324
so how do we actually know 3a^2 + 3b^2 = 35?


Just to be clear, this is wrong. DFranklin probably didn't bother seeing where you got this from so he didn't pick up on it for you.

What you wrote here should be 3a2+3b2=−q3a^2+3b^2 = -q because the LHS is literally the product of all roots: 3(a+ib)(a−ib)3(a+ib)(a-ib).
Original post by RDKGames
Just to be clear, this is wrong. DFranklin probably didn't bother seeing where you got this from so he didn't pick up on it for you.

What you wrote here should be 3a2+3b2=−q3a^2+3b^2 = -q because the LHS is literally the product of all roots: 3(a+ib)(a−ib)3(a+ib)(a-ib).

ok so i got alpha as -2, 3--2 =5, so 5b = 35 and b = 7, meaning the roots are -2+7i and -2-7i ? im doing the sigma alpha beta stuff and i seem to be getting the right answers now so thank you for helping, unfortunately still stuck on the one with w^2, is there a way to find intersections between an arguement and a circle?
Original post by Gent2324
ok so i got alpha as -2, 3--2 =5, so 5b = 35 and b = 7, meaning the roots are -2+7i and -2-7i ? im doing the sigma alpha beta stuff and i seem to be getting the right answers now so thank you for helping


That's right, but don't forget to actually answer the question. It is not asking for the roots.

unfortunately still stuck on the one with w^2, is there a way to find intersections between an arguement and a circle?


If we denote w=a+ibw = a + ib then ∣w∣2=a2+b2|w|^2 = a^2 + b^2


Then ∣w−1−i∣=∣(a−1)+i(b−1)∣=(a−1)2+(b−1)2=3|w - 1 - i| = |(a-1) + i(b-1)| = \sqrt{(a-1)^2 + (b-1)^2} = 3

and also arg⁡(w−2)=arg⁡[(a−2)+ib]=π4\arg(w-2) = \arg[(a-2)+ib]= \dfrac{\pi}{4} here means arctan⁡ba−2=π4\arctan \dfrac{b}{a-2} = \dfrac{\pi}{4}.


So you can solve for a,ba,b and sub them into a2+b2a^2 + b^2.
(edited 5 years ago)
Original post by RDKGames
That's right, but don't forget to actually answer the question. It is not asking for the roots.



If we denote w=a+ibw = a + ib then ∣w∣2=a2+b2|w|^2 = a^2 + b^2


Then ∣w−1−i∣=∣(a−1)+i(b−1)∣=(a−1)2+(b−1)2=3|w - 1 - i| = |(a-1) + i(b-1)| = \sqrt{(a-1)^2 + (b-1)^2} = 3

and also arg⁡(w−2)=arg⁡[(a−2)+ib]=π4\arg(w-2) = \arg[(a-2)+ib]= \dfrac{\pi}{4} here means arctan⁡ba−2=π4\arctan \dfrac{b}{a-2} = \dfrac{\pi}{4}.


So you can solve for a,ba,b and sub them into a2+b2a^2 + b^2.

im lost as to where u got (a-1+i(b-1) from? and the third line too
by replacing w with a+bi you get a+bi-1-i
Original post by Gent2324
im lost as to where u got (a-1+i(b-1) from? and the third line too


w−1−i=(a+ib)−1−i=a−1+ib−i=(a−1)+i(b−1)w-1 - i = (a+ib) - 1 - i = a-1 + ib - i = (a-1) + i(b-1)

Likewise for w−2=(a−2)+ibw-2 = (a-2)+ib.


If you're confused on where I get arctan⁡ba−2=π4\arctan \dfrac{b}{a-2} = \dfrac{\pi}{4}, then have a think about how you would determine e.g. arg⁡(1+2i)\arg (1+2i) and compare this with the result I've provided. It should make sense.
(edited 5 years ago)
Original post by RDKGames
w−1−i=(a+ib)−1−i=a−1+ib−i=(a−1)+i(b−1)w-1 - i = (a+ib) - 1 - i = a-1 + ib - i = (a-1) + i(b-1)

Likewise for w−2=(a−2)+ibw-2 = (a-2)+ib.


If you're confused on where I get arctan⁡ba−2=π4\arctan \dfrac{b}{a-2} = \dfrac{\pi}{4}, then have a think about how you would determine e.g. arg⁡(1+2i)\arg (1+2i) and compare this with the result I've provided. It should make sense.

oh right so youre turning the w-1-i into cartesian form?
after that though im still confused, i kind of get the arctan bit, meaning b^2 + (a-2)^2 = w^2 ? which is a + b = w, after that im lost though, like i dont get why w-2 = (a-2) + ib
Original post by Gent2324
oh right so youre turning the w-1-i into cartesian form?
after that though im still confused, i kind of get the arctan bit, meaning b^2 + (a-2)^2 = w^2 ? which is a + b = w, after that im lost though, like i dont get why w-2 = (a-2) + ib


What are you on about here?

b2+(a−2)2=w2b^2 + (a-2)^2 = w^2 is wrong.
a+b=wa+b = w is wrong. (Come on, we defined w=a+ibw = a+ib at the beginning!)

And how do you still not get w−2=(a−2)+ibw-2 = (a-2)+ib ?? We have w−2=(a+ib)−2=a−2+ib=(a−2)+ibw-2 = (a+ib)-2 = a-2 + ib = (a-2) + ib


All I am doing here is replacing ww everywhere and separating the expressions into clear x+iyx+iy Cartesian forms.


Anyway, arg⁡ba−2=π4\arg \dfrac{b}{a-2} = \dfrac{\pi}{4} so b=a−2b=a-2. Sub that into (a−1)2+(b−1)2=9(a-1)^2 + (b-1)^2 = 9 and solve for the solution where a>2a > 2 since the intersection clearly happens to the right of z=2z=2 (look at your sketch from part a!)
Original post by RDKGames
What are you on about here?

b2+(a−2)2=w2b^2 + (a-2)^2 = w^2 is wrong.
a+b=wa+b = w is wrong. (Come on, we defined w=a+ibw = a+ib at the beginning!)

And how do you still not get w−2=(a−2)+ibw-2 = (a-2)+ib ?? We have w−2=(a+ib)−2=a−2+ib=(a−2)+ibw-2 = (a+ib)-2 = a-2 + ib = (a-2) + ib


All I am doing here is replacing ww everywhere and separating the expressions into clear x+iyx+iy Cartesian forms.


Anyway, arg⁡ba−2=π4\arg \dfrac{b}{a-2} = \dfrac{\pi}{4} so b=a−2b=a-2. Sub that into (a−1)2+(b−1)2=9(a-1)^2 + (b-1)^2 = 9 and solve for the solution where a>2a > 2 since the intersection clearly happens to the right of z=2z=2 (look at your sketch from part a!)

ohhh i think i see now, since the the line made by the arg is just y=x-2, and then you just factor that in as the intersection with the loci. thank you i see it now

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