4U Maths Question Thread

[flacko]

Hey been struggling with 2 questions from Patel.

"For the following, describe the locus of the complex number w, where z is restricted as indicated.

w=(z-2+i)/(z+2-i) , |z|=1

Thanks

[hassrax]

Hi i need help with this q, if z is a complex number, z/z-i  then show that it is imaginary. I let z=x+iy and tried doing it from there but i get stuck when i let re(z) =0 , thanks

[terassy]

Help please 

P and Q are variable points on the rectangular hyperbola xy = c². The tangents at P and Q meet at R. If PQ passes through the point (a,0), find the equation of the locus of R.

Okay,
So I've got the equation of the tangent at P:
x=2cp-p²y

and the tangent at Q
x=2cq-q²y

Simultaneously solve them and R is at:
x= 2cp/(p+q)
y=2c/(p+q)

And the equation of the chord PQ is:
x+pqy=c(p+q), but it passes through (a,0), so we get this:
a=c(p+q)

Now what do I do. I would usually try to form an equation by getting rid of the parameters (p+q), but then pq is left behind so what do I do?

[jamonwindeyer]

Hi i need help with this q, if z is a complex number, z/z-i  then show that it is imaginary. I let z=x+iy and tried doing it from there but i get stuck when i let re(z) =0 , thanks

Hey! Super sorry about the delay in response - I'm not sure if everything is quite right with this question. Are we trying to prove that a complex number of the form \(\frac{z}{z-i}\) is imaginary for any \(z\)? That isn't quite true. If we're trying to prove that the solution to \(z=\frac{z}{z-i}\) is imaginary then that isn't quite true either (answers are \(z=0\) and \(z=1+i\)). Maybe I'm misinterpreting you too - Could you maybe share the question/source?

[jamonwindeyer]

Help please 

P and Q are variable points on the rectangular hyperbola xy = c². The tangents at P and Q meet at R. If PQ passes through the point (a,0), find the equation of the locus of R.

Okay,
So I've got the equation of the tangent at P:
x=2cp-p²y

and the tangent at Q
x=2cq-q²y

Simultaneously solve them and R is at:
x= 2cp/(p+q)
y=2c/(p+q)

And the equation of the chord PQ is:
x+pqy=c(p+q), but it passes through (a,0), so we get this:
a=c(p+q)

Now what do I do. I would usually try to form an equation by getting rid of the parameters (p+q), but then pq is left behind so what do I do?

Hey! So we're here for the coordinates of R (think you forgot a \(q\) in your message?):



And as you correctly get to, \(p+q=\frac{a}{c}\), so we can ditch that. Where do we ditch \(pq\)? So it's a bit of a trick - We don't need to! Take the y-coordinate:



That is the locus! It's a horizontal line, you don't need to bring the \(x\) at all. Our objective was to get an equation with no \(p\) or \(q\), and we have that here. The locus is the horizontal line \(y=\frac{2c^2}{a}\)

Watch out for this in locus questions, sometimes you don't need to bring the \(x\) in

[Livjane_2203]

Would appreciate some help please.

Use De Moivre's Theorem to solve the equation z^5=1. Show that the points representing the five roots of this equation on an Argand diagram form the vertices of a regular pentagon of area 5/2 sin 2pi/5 and perimeter 10 sin pi/5.

I understand how how to get the roots but not sure how to get to the area and perimeter.

Thanks. 

[jamonwindeyer]

I understand how how to get the roots but not sure how to get to the area and perimeter.

Hey there! Nice job, you've done the hard bit! For the area and perimeter, pick two of the roots you found and the origin on the complex plane. It forms a triangle, and that triangle is the basis on which we'll get the area/perimeter.

- We can find the area of the triangle using \(A=\frac{1}{2}ab\sin{C}\), where \(a\) and \(b\) are the moduli of the two  roots (they'll be the same), and \(C\) is the angle at the origin (it is \(\frac{2\pi}{5}\), since it is a fifth of the full angle at the middle). Multiply that area by five to get the area of the pentagon.
- We can find the length of the side opposite the origin by using the cosine rule, same lengths/angles as above. Multiply that by five to get the perimeter.

There are definitely other ways, this is a geometrical approach which I think is pretty simple hope this helps!

[david.wang28]

Hello, just a question on 4U Cambridge, can anyone please give me a detailed answer for question 7 and 8 in the link? Thanks

[Divayth Fyr]

Hello, just a question on 4U Cambridge, can anyone please give me a detailed answer for question 7 and 8 in the link? Thanks

Ah, these questions. I remember when I first saw them; it was not a pleasant discovery. I hope the solutions help.

[g.xzhu]

Hi,

Could I please get some help with these two maths induction questions?

For all positive intergers n, prove by maths induction that:
     1) |z₁z₂...z_n| = |z₁| |z₂| ... |z_n|
     2) arg(z₁z₂...z_n) = arg(z₁) + arg(z₂) + ... + arg(z_n)

Thank you!

[david.wang28]

Ah, these questions. I remember when I first saw them; it was not a pleasant discovery. I hope the solutions help.

Thanks for the help!

[jamonwindeyer]

Hi,

Could I please get some help with these two maths induction questions?

For all positive intergers n, prove by maths induction that:
     1) |z₁z₂...z_n| = |z₁| |z₂| ... |z_n|
     2) arg(z₁z₂...z_n) = arg(z₁) + arg(z₂) + ... + arg(z_n)

Thank you!

Hey! Let's roll with (1), I'll assume you could prove it for \(n=1\) (it is self apparent), but we'll also want to prove it for \(n=2\) to use later.



We can prove this by letting \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\), and substituting.



Find each of those moduli (expand out the LHS) and they will be the same. So you've proved the case for \(n=2\).

Now, the induction assumption for (1) is:



Now let's go with \(n=k+1\) and see what we get:



Let's break this into two pieces, \(a=z_1z_2...z_k\) and \(b=z_{k+1}\). We do this because we've already proven that \(|z_1z_2|=|z_1||z_2|\) above! So by splitting it in two, we can automatically therefore say that:



Now we use our induction assumption!!



And we are done!! Conclude as usual, and you're all set the second question is the exact same process, use the result for \(n=2\) to help you generalise it for \(n=k+1\)

[g.xzhu]

Hey! Let's roll with (1), I'll assume you could prove it for \(n=1\) (it is self apparent), but we'll also want to prove it for \(n=2\) to use later.



We can prove this by letting \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\), and substituting.



Find each of those moduli (expand out the LHS) and they will be the same. So you've proved the case for \(n=2\).

Now, the induction assumption for (1) is:



Now let's go with \(n=k+1\) and see what we get:



Let's break this into two pieces, \(a=z_1z_2...z_k\) and \(b=z_{k+1}\). We do this because we've already proven that \(|z_1z_2|=|z_1||z_2|\) above! So by splitting it in two, we can automatically therefore say that:



Now we use our induction assumption!!



And we are done!! Conclude as usual, and you're all set the second question is the exact same process, use the result for \(n=2\) to help you generalise it for \(n=k+1\)

Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?

Thanks!

[jamonwindeyer]

Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?

Thanks!

Hey! Nicely done - Definitely looks cumbersome, but nothing immediately jumps to me as easier. If anyone reading this spots anything, give a shout

[RuiAce]

Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?

Thanks!

Hey! Nicely done - Definitely looks cumbersome, but nothing immediately jumps to me as easier. If anyone reading this spots anything, give a shout

The quicker way was to let \(z = r \operatorname{cis} \theta\) instead of go for the Cartesian approach. That reduces the working out by a tiny bit for the modulus, but makes the argument just as fast as the modulus to deal with