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It's been a while since I did this topic. Dunno if I'm right, but hoping to help you at least. As Liam said, since there must be at least 4 balls of each colour; 4x5=20, gives us 20 balls fixed in place. Hence, we have to find the remaining combinations. Since 1 colour can be no more than...

It does make sense, and I'm thankful for the clarity. I struggle with these subtleties, but I think they are very important and try and have them sink in whenever I come across them.

Interesting... Is the equation Arg(z-i) = Arg(z) - pi ? I had thought +pi or -pi didn't matter because they both ended up at the same place. Since the Complex plane is restricted between -pi and pi, I thought a result of 3pi/2 simply means the same as -pi/2. I simply saw +pi as a rotation...

I re-did it and went the exact same route as you. I agree with you 100% now. Thx for putting up with me 🤡🤡🤡

yeah, i completely misread it. My bad

What's the equation for (A)? o_O

oh wait, i see how you are interpretting it. You are reading it as: |v| = root(2) |w| Where as, I'm reading it as |u| = root 2 |v| = root 3 |w| = root 3 It's ambiguous from the OP. (or im dumb. yeah, maybe im dumb haha)

where does the questions say |v| = root (3) |w| The questions says |w| AND |v| = root(3) which means they are both equal to root (3)

If |v| = √3, how can |v| = 3 ? (bolded) Also bolded: |w| and |v| = √3. Doesn't this mean |w| = √3 and |v| = √3 ? Otherwise, what does it mean?

(|u|^2) / (2) = (2) / (2 |u| |v| ) = 1 / (|u| |v|) --- Oops... That means my pi - theta is actually theta. So cos theta = root 6 / 6 And by constructing a right triangle: sin theta = root (30) / 6 which is still the same final answer (somehow). New working is attached as Q00001i.jpg ---...

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Sweet as. Please confirm my answers The equation for each graph: A. Arg(z-i) = Arg(z)+pi B. Arg(z-i) = Arg(z) C. Arg(z) = Arg(i) D. |z| = |z-i|

The locus equation should be y=x-1 (not y=x+1), right? Your algebra checks out, but it doesn't give the correct equation of the locus. My consideration for what's problematic would be the value of inverse tan as it isn't clear what quadrant each complex number is in.

Trying to solve algebraically feels garbage. (or I'm garbage. one or the other) The visual explanation makes it very easy though.

Here's a question and solution. Originally, I only considered when arg(z) is +ve (or when Im(z) > 0). After checking the answers and some thought, here is my solution. My way of thinking is sufficient to answer the question, however, it feels strange. I was wondering if anyone has any...

Hope it helps.

Got it. Is this year 10 circle geometry? Got it! thankyou. too easy. I'm embarrassed I didn't see this one, thanks for your time. I constructed triangles by connecting the different colours to red, however, it's difficult to tell which angle is the largest. The circles with a touching...

Here's my worked solution.

Hi guys, I have a few questions about this chapter. Please help, if you can. And if you have questions of your own from this chapter of the cambridge book, pls post them here and I'll try help as well. My questions: 1F Q12.a) After drawing the graph and looking at the answers, it's clear that...