Author Topic: A.level Pure 3 MATH - HELP!!  (Read 1126 times)

Offline Tammet

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A.level Pure 3 MATH - HELP!!
« on: May 19, 2010, 08:18:36 am »
Can someone please tell me how to draw |z+4| = 3|z| on an Argand diagram.. with detailed solution please.

Thanks.

Offline astarmathsandphysics

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Re: A.level Pure 3 MATH - HELP!!
« Reply #1 on: May 19, 2010, 09:45:54 am »
Square both sides to get |z+4|^2=9|z|^2. Put z=x+iy then |z+4|^2=|x+4+iy|^2=(x+4)^2+y^2 and 3|z|^2=3x^2+3y^2 put these equal and simplify to get 2x^2-8x+2y^2-16=0 which is a circle

Offline Tammet

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Re: A.level Pure 3 MATH - HELP!!
« Reply #2 on: May 19, 2010, 04:37:57 pm »
That's what I exactly did before, and i ended up with that equation.
But I thought all my working was wrong because, how is that an equation of
a circle? what's the center and the radius? Shouldn't we complete the square?

Offline astarmathsandphysics

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Re: A.level Pure 3 MATH - HELP!!
« Reply #3 on: May 19, 2010, 10:33:18 pm »
2x^2-8x+2y^2-16=0
divide by 2
x^2-4x+y^2 -8=0
(x-2)^2 -4+y^2 -8=0
(x-2)^2 +y^2 =12
centre (2,0) radius sqrt(12)

Offline Tammet

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Re: A.level Pure 3 MATH - HELP!!
« Reply #4 on: May 20, 2010, 03:20:54 am »
Uhmm thanks for your reply man. But my book says the answer is

"Circle center (0.5+2i), radius=1.5. Circle equation: x(squared)+y(squared)-x=2"

any idea? =/

Offline astarmathsandphysics

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Re: A.level Pure 3 MATH - HELP!!
« Reply #5 on: May 20, 2010, 07:51:09 am »
No way you can get centre at 0.5+2i cos there is no y term. Mistake in ms

Offline Tammet

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Re: A.level Pure 3 MATH - HELP!!
« Reply #6 on: May 20, 2010, 12:15:33 pm »
I guess so.
Thanks anyways.