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if the complex number Z lies on the circle |Z-2i| = 2sqrt(2) then find arg( (Z-2)/Z+2) ) ?
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Omg.. Hate complex no. !
Unfortunately, I love complex no. :( :)
Carry on bro ! :D D:
Did I say that I know how to solve this question? :P , please wait, I am trying
@Yahoo! try to put : z = x + iy
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then continue.
|Z-2i| = 2sqrt(2) is a circle with centre (0,2) and radius 2sqrt(2)///
and i think we have to do something with that..:)
I can say that : \(x^2 - y^2 + 4y = 12\)
\(x+iy = Z\)
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|dw:1366255497050:dw|
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LoL It's you @jamespatrick1071
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\[|Z-2i| = 2\sqrt{2}\] |dw:1366263056433:dw| \[arg( \frac{Z-2}{Z+2} )=arg(z-2)-arg(z+2)\]
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