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November 01, 2025, 04:20:50 pm

Author Topic: Tea.towel's Specialist Questions  (Read 854 times)  Share 

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tea.squaredd

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Tea.towel's Specialist Questions
« on: January 25, 2011, 11:09:54 pm »
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Hi guys,

I'm taking specialist as a subject this year and there's bound to be questions that I'll get stuck on. So please help me guys :D !


Cheers!

Here's the 1st question:

z= 9acos(theta) + 4asin(theta)i, Find |z|

Sorry i dont know how to do those actual math signs on the forum yet.. !
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evaever

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Re: Tea.towel's Specialist Questions
« Reply #1 on: January 26, 2011, 12:13:07 am »
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|z|= sqrt((9acos(theta))^2 + (4asin(theta))^2), squaring whatever

tea.squaredd

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Re: Tea.towel's Specialist Questions
« Reply #2 on: January 26, 2011, 01:32:42 am »
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Really? So the answer would be..
sqrt [81a^2cos^2(theta) + 16a^2sin^2(theta) ]
S:
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david10d

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Re: Tea.towel's Specialist Questions
« Reply #3 on: January 26, 2011, 01:48:07 am »
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disregard


« Last Edit: January 26, 2011, 02:13:42 am by shaGteM »
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evaever

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Re: Tea.towel's Specialist Questions
« Reply #4 on: January 26, 2011, 02:04:13 am »
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you forgot i^2 = -1 which makes it:

sqrt [81a^2cos^2(theta) - 16a^2sin^2(theta)]

which then you can further factorise to:

sqrt [(9acos(theta)+4asin(theta)(9acos(theta)-4asin(thetic)]

are you sure that there's nothing left to the question? it just seems out of place lol




z=x+iy, |z|=Sqrt(x^2+y^2) NOT Sqrt(x^2-y^2)

Dr.Lecter

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Re: Tea.towel's Specialist Questions
« Reply #5 on: January 26, 2011, 02:07:55 am »
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you forgot i^2 = -1 which makes it:

sqrt [81a^2cos^2(theta) - 16a^2sin^2(theta)]

which then you can further factorise to:

sqrt [(9acos(theta)+4asin(theta)(9acos(theta)-4asin(thetic)]

are you sure that there's nothing left to the question? it just seems out of place lol




z=x+iy, |z|=Sqrt(x^2+y^2) NOT Sqrt(x^2-y^2)

wtf wrong
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david10d

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Re: Tea.towel's Specialist Questions
« Reply #6 on: January 26, 2011, 02:13:10 am »
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oh shit. lol yeah my bad... relax it's the wee hours of the morning

now it seems that a has to equal to something to make the identity sin^2(A)+cos^2(A)=1
« Last Edit: January 26, 2011, 02:20:38 am by shaGteM »
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pHysiX

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Re: Tea.towel's Specialist Questions
« Reply #7 on: February 09, 2011, 08:44:00 am »
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Now you can keep that answer or use the identity to make |z| entirely cos or sin :D

or

To learn the maths signs, use [t e x] <insert code> [/t e x] wrapper (without the spaces there).
For the code: http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/  :)
« Last Edit: February 09, 2011, 09:02:18 am by pHysiX »
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