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complex no (1 Viewer)

martin310015

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can some one help me out

let w= cos8+isin8 and z=w+1/w-1

show that z=-i cot8 8 = theta

Thanx
 

wogboy

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Tip: Use @ for theta rather than the number 8.

Are you trying to prove
w = cos@ + isin@ and z = w + 1/w - 1,
so then z = -icot@ ?

It's not true, e.g:

let @ = pi/2,
then w = i
so z = i + 1/i - 1 = -1

but -icot@ = 0

-1 =/= 0
 

martin310015

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what i got is icot @ from using the half angles.
 
Last edited:

:: ck ::

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100bux says u go to prior

questoin 7a from the homework :) stop cheating

btw theres a correction its supposed to be 2@
double angle :)
 

ae

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i think the question suppose to read
"w = cos2@ + isin2@ and z = w + 1/w - 1"

then its easy to prove z=-icot@
 

Mill

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$200 says I'm his class tutor, stop cheating Martin L :)
 

Mill

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Actually I put the question in the worksheet.

Originally it was meant to be w = cos@ + isin@, but I decided that was to hard to simplify / work out (the answer in that case would be -icot[@/2] ). I guess it just wasn't fixed up in the homework sheet.

There's an alternative method of solving it, but its much more boring. You just realise the denominator. It's just a straightforward algebra bash that way.
 

martin310015

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hahahaha
its not really cheating :)
no one posted the working outs

its the workin out that counts right...
 

martin310015

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but shouldn't the answer be positive icot@/2 ........
for w=cos@+isin@
 

Mill

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Yeah for the question written in the actual homework the answer you should get is -icot@/2. But the question written in the homework was wrong.

It should have been w=cos2@+isin2@, to give the answer of -icot@.

I'm fairly sure I changed it in your book, but marked it right anyway since you got -icot@/2 if I'm not mistaken.
 

Grey Council

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So Mill's is a tutor at Prior. mmm

How about you post up some questions, Mills? :)
 

Mill

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Haha the qood questions I've found or created in the last few months have found their way into Prior worksheets, so I gots nothing for ya!
 

martin310015

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i got a question
use mathematical induction to prove that the |z^n|=|z|^n

for this question how would i start it and do i use the modulus argument form or x+iy
 

shkspeare

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haha and u kno how much prior sheets are worth guardian? lolz i dun think ppl wud like it if it was given out :p
 

Mill

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I have 1 workshop haha. I didn't want any, and I think as soon as your 4u class gets more people I won't have the workshop anymore. My 1 workshop is at the end of your class (like during the end of it) so you won't be able to be in it.
 

Mill

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Just looking at it I'd try mod-arg form first since if z = rcis@, then |z| = r, which is nice and simple. I'll check it now though.
 

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