complex number problem (2 Viewers)

[Dash]

Maybe he just didn't want the extra workload and was happy with 3u...
I mean, that probably would have been my reason for not doing it in his situation.

 

[abdooooo!!!]

no... didn't you read my post people... he wants to top 2u now.

because he can do 4u next year or whatever so he can top it later... ahahah.

edit: sorry AGB.

 

[Giant Lobster]

Hey whats all this talk on topping 3u, im confused. Are ext2 students excluded when they deduce who topped 3u? Or does it include ext2 students? if it was the latter then statistically ext2 students are much more likelier to top 3u hmm

 

[abdooooo!!!]

i don't know... i think now its the latter. still im not sure... but i wonder why no one top 4u, top 3u.

statistically this should be a common occurence... LOL.

umm... im gonna practice heaps of 3u now if 4u student can top it, so i don't make any mistakes and hope that the test is easy and i score 84/84... hahaha.

 

[Xayma]

Originally posted by abdooooo!!!
i don't know... i think now its the latter. still im not sure... but i wonder why no one top 4u, top 3u.

Well we had a top 4U, second in 3U so its probably just due to stupid mistakes.

On the same line as that, say a 3U person and a 4U person score the same on the ext 1 test.

Say that this score gives the 4U person an odd score after aligning, would the 3U's person (whose would only be out of 50) have 1/2 rounded up or down?

Eg. The 4U person got 87 Aligned, would the 3U person get 43 or 44

 

[KeypadSDM]

Originally posted by Xayma
Well we had a top 4U, second in 3U so its probably just due to stupid mistakes.

It was stupid mistakes.

Anyway, top of 3 unit did 4 unit. Nuff said.

 

[siavash_s_s]

hey abdooo why not go into modarg form? makes it alot easier imo, no sqrt0.5 etc..., just pi (gotta love pi)

 

[victorling]

Originally posted by AGB
i know that this is a really easy question, but i just dont get it...

find all solutions to z^4 = -1

can someone please explain this to me??

(1)change -1 to mod-arg form----->cis(180)
(2)Add 2(180)k to it--------->cis(180+360k)
(3)Use De moviem's theorem
(4)solve and sub k=0,1,2,3