Complex numbers question (1 Viewer)

vds700 · Sep 27, 2008

(b) The complex number β is given by β = (1 − t<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w

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(i) Show that |β| = 1+t<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w

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ontGrowAutofit/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><style>  </style>² and arg β = θ where t = tanθ/2. I/m fine with this part.

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(ii) Write down the modulus and argument of one square root of β, and hence if
z<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w

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conics2008 · Sep 27, 2008

Hey ill give the quick steps, but not the full working out and I hope you can continue on from there.

z^2= B where B = (1-t^2)+(2t)i

Let z= a+ib

therefore (a+ib)^2= 1-t^2 +2ti { Expand LHS and Equate real and Imginary }

a^2-b^2= 1-t^2 (1) and 2t = 2ab (2)

from the imginary parts we can conclude that a=t/b and therefore sub this into (1)

{t^2/b^2) - b^2 = 1-t^2 simplify this equation then let m=b^2

after that you should get an equation like this

m^2 +m(1-t^2) -t^2 =0

Now use quad equation

-(1-t^2) +- root of (1-t^2)-4(-t^2) /2

expand and simpfly the square root you should get at the end

-1+t^2+-(1+t^2) /2

now this will give you the two roots.. when you find them just apply part i on them and you shall get your argument and modulus.

=] hope that helps.. if you still having trouble just let me know and ill post up full solution.

Good luck for the hsc

tanguyen · Sep 27, 2008

Hmm... someone correct me if I'm on the wrong track, its been a year since I've visited these boards.

You've already worked out the modulus and argument in part (i) so you pretty much have the complex number (Beta) in mod-arg form (fancy that). You can easily take the square root of that by square rooting the modulus (the question asked for just one square root, so you can either take a positive or negative number for the square root of the modulus) and utilising DeMoivre's for the argument.

If z^2 = (beta), then z = (plus/minus)*(Beta)^(1/2). You've already found one root of (squareroot)Beta so its just a matter of expanding the mod-arg form out to get the Cartesian form. How does that sound?

vds700 · Sep 27, 2008

tanguyen said:
Hmm... someone correct me if I'm on the wrong track, its been a year since I've visited these boards.

You've already worked out the modulus and argument in part (i) so you pretty much have the complex number (Beta) in mod-arg form (fancy that). You can easily take the square root of that by square rooting the modulus (the question asked for just one square root, so you can either take a positive or negative number for the square root of the modulus) and utilising DeMoivre's for the argument.

If z^2 = (beta), then z = (plus/minus)*(Beta)^(1/2). You've already found one root of (squareroot)Beta so its just a matter of expanding the mod-arg form out to get the Cartesian form. How does that sound?

yeah well β = (1+t^2)cisθ so sqrt β=sqrt[1 + t^2)cisθ] so the modulus is sqrt(1 + t^2), but how do u find the argument? I thought De Moivre's theorem ony applied for integers.

Maybe conics' way is the best method, though from the question, i get the impression they want you to fing the modulus and argument firs, then find z.
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Complex numbers question (1 Viewer)

vds700

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conics2008

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tanguyen

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vds700

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