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ASAP: I have to identify a series for the number of handshakes that can be made for 'n' people. Is the below series right? 0 + …. + (n-2) + (n-1). I have had some people tell me that it is 0 + 1 + ... + (n-1), but I believe that a group with one person can make zero handshakes - so what would...

I have got a multimodal coming up and am a bit confused as to how I could personally engage with texts... any idea on how to do this?

@CM_Tutor yes I agree - it was just that I got a complex number that was purely real when simplified and I wrote the real part and forgot to write 0i as I thought that it was not needed.

@Drongoski oh ok! Thank you for reassuring!

Will you lose a mark if you write an imaginary with 0 imaginary part as just the real part? I might have made that mistake in my first test, so I just wanna kind of know...

Also can someone show me how to do part (5)?

@5uckerberg was it his roots. One his roots were supposed to be e^i(-4pi/5) I believe.... @ExtremelyBoredUser thanks btw!

How would you do part 5 of this question? I have done questions like these before, but somehow can't find the method I used before. Thanks!

I guess I can just use sheer brute force to sketch the following curves (especially b, c and d). But is there a more smarter way of approaching such types of questions???

I observed that for purely imaginary number cases, it is generally something times (k + 1) set of values - is this the case for most problems?

@5uckerberg and @ExtremelyBoredUser fair. I guess I will just show the entire working out based on the mark allocation, just to stay on the safe side!!!

For questions like above, do we need a lot of steps to show that something is true. I mean it felt it was obvious. How should I approach this?

Ah thank so much you! @5uckerberg

correct me if I am wrong, both the constituents of alpha and omega are conjugates right. So can I just use (z^2 -2Re(z)z + 1) and represent it as quadratics that way?

How would you do (b) in particular?

@Vipul_K Just plug in x = iy and then solve for o. In part (a) you solve for the imaginary part to be zero and then sketch that graph and in part (b) you solve for the real part to be zero and then sketch that graph. Make sure to exclude any points by looking at the domain of the graph...

oh ok. could you please kinda elaborate on the positive and negative numbers part?

I do know that you split up the arguments as arg(z-2) = arg(z) for part (a). But can someone please explain generally the reasoning and methodology behind answering this type of questions? Thanks!

@5uckerberg Yea true I saw this elsewhere as well!

@5uckerberg That makes total sense now - thank you!