4U Maths Question Thread

[frog1944]

for any complex numbers z1,z2, show that |z1+z2|^2 + |z1-z2|^2 = 2(|z1|^2 + |z2|^2)
thank you

An alternative method, without resorting to the modulus-argument form is;

[RuiAce]

2i(y-1)>0

Can you divide by 2i and keep the inequality the same?

To give y>1

Or is there something I'm missng?

Inequalities don't exist with complex numbers (because they don't make sense). You can do inequalities with the real/imag parts, the mod and the arg, but not the complex number itself. The fact that you got there means you most likely made a mistake along the way.

An alternative method, without resorting to the modulus-argument form is;

Lol right, that's what I failed to think up last night. I knew the linear algebra way but entirely forgot the complex analysis version.

@sssona09 please use this method instead

[maria.micale]

Hey is there a site where is can plug in things including arg and it will sketch it on complex plane? My textbook docent have answers to some of the questions.

I've used the complex plane on desmos but I can't or don't know how to type in arg.

If not then can you sketch for me:

 -π < arg < π

 arg (z - i) = 0

arg ( z + 2) = 3π/4

[RuiAce]

Hey is there a site where is can plug in things including arg and it will sketch it on complex plane? My textbook docent have answers to some of the questions.

I've used the complex plane on desmos but I can't or don't know how to type in arg.

If not then can you sketch for me:

 -π < arg < π

 arg (z - i) = 0

arg ( z + 2) = 3π/4

If you want to use technology, consider feeding GeoGebra the following input:

Code: [Select]

arg(x+i*y+2)=3pi/4
Note that excluding -2 will be necessary. GeoGebra won't do this, but you will need to.

[maria.micale]

If you want to use technology, consider feeding GeoGebra the following input:

Code: [Select]

arg(x+i*y+2)=3pi/4
Note that excluding -2 will be necessary. GeoGebra won't do this, but you will need to.

Thanks Rui,

Could you please sketch for me -π < arg < π because I don't think geogebra can do it?
Or if not, would you sketch the whole plane excluding the real axis ( i = 0)?

[RuiAce]

Thanks Rui,

Could you please sketch for me -π < arg < π because I don't think geogebra can do it?
Or if not, would you sketch the whole plane excluding the real axis ( i = 0)?

Well yeah -π < arg < π is essentially the whole plane.

But you would exclude only the negative real axis, and the point 0. Because anything on the positive real axis satisfies arg(z) = 0, which still lies between -π and π. Only those on the negative real axis satisfy arg(z) = π, which just barely lies outside that range.

[Caleb Campion]

I just don't know where to start when attacking this question? Is sum of roots the right way to go? Would love if I could get a worked solution

cos(pi/7) = cos(2pi/7) + cos(4pi/7) + 1/2

[RuiAce]

I just don't know where to start when attacking this question? Is sum of roots the right way to go? Would love if I could get a worked solution

cos(pi/7) = cos(2pi/7) + cos(4pi/7) + 1/2

Sum of roots will indeed be necessary but you would certainly not be told to prove this in an exam without any guidance. Please provide the source of the question.

[maria.micale]

Hey how would you go about factorising z^5 + 3z^4 - z - 3?

I got the first 3 factors by finding factors of the constant term that sub into P(z) to get zero.

So far I got P(z) = (z-1)(z+1)(z+3)Q(z)

But I don't know the best way to go about finding Q(x).

[Eric11267]

Hey how would you go about factorising z^5 + 3z^4 - z - 3?

I got the first 3 factors by finding factors of the constant term that sub into P(z) to get zero.

So far I got P(z) = (z-1)(z+1)(z+3)Q(z)

But I don't know the best way to go about finding Q(x).

I would do it by grouping terms

Edit: I don't know if this is complex numbers or whatever so I'll just leave it like this

[maria.micale]

I would do it by grouping terms

Edit: I don't know if this is complex numbers or whatever so I'll just leave it like this

Thankyou! Also,

How do I factorise this?

z^4 + 4z^3 + 3z^2 - 8z - 10 if I am given (z +2 - i) as one linear factor.

I know that (z -2 + i) would be another factor due to a theorem but I don't know what to do to get the other two factors.

[Sine]

Thankyou! Also,

How do I factorise this?

z^4 + 4z^3 + 3z^2 - 8z - 10 if I am given (z +2 - i) as one linear factor.

I know that (z -2 + i) would be another factor due to a theorem but I don't know what to do to get the other two factors.

If (z + 2 -i) is a factor then (z + 2 +i) is also a factor via the conjugate root theorem (probably a typo on your part)
From here I would expand the two factors that you have i.e. expand (z + 2 -i)(z + 2 +i) which would yield a real quadratic.From here you can long divide the quartic that you have with the quadratic to yield another quadratic then we can factorise this final quadratic using the techniques we know.

[KT Nyunt]

Hey 

I have a 4U assessment coming up and I have this question:

If given the graph of f(x), how would I sketch f^2(x)?

Is this the same as f(f(x))?


Thanks in advance

[RuiAce]

Hey 

I have a 4U assessment coming up and I have this question:

If given the graph of f(x), how would I sketch f^2(x)?

Is this the same as f(f(x))?


Thanks in advance

You should consider what the source of your question has done and follow their rule.

[maria.micale]

Hi, Could someone please explain how to factorise 2z^2 - 2z +1 over C?

I used quadratic formula so that z = 1/2 + or - i/2
So I thought P(z) = (z - 1/2 + i/2)(z - 1/2 - i/2)

But the answer is P(z) = 2(z - 1/2 + i/2)(z - 1/2 - i/2) (s0 there is a 2 out the front, I thought this may have something to do with the polynomial being non-monic but I'm not sure)