Complex no help (1 Viewer)

darkphoenix · Dec 2, 2011

ok, so i was revising complex no. and one question asks to find the locus of w= z-2i/1-z absolute value (z) =2
and the answer is a circle with radius: root 20/3 and center (-4/3, 2/3)
I don't get why my working doesn't work:
[] = absolute value
z-zw = z-2i
z(1+w)=w+2i
z= w+2i/1+w
2= [(w+2i)/(1+w)]
2[w+2i] = [1+w]
and use the distance formula to find the equation

Can someone explain with solutions to this, thanks

RealiseNothing · Dec 2, 2011

darkphoenix said:
ok, so i was revising complex no. and one question asks to find the locus of w= z-2i/1-z absolute value (z) =2
and the answer is a circle with radius: root 20/3 and center (-4/3, 2/3)
I don't get why my working doesn't work:
[] = absolute value
z-zw = z-2i
z(1+w)=w+2i
z= w+2i/1+w
2= [(w+2i)/(1+w)]
2[w+2i] = [1+w]
and use the distance formula to find the equation

Can someone explain with solutions to this, thanks

The LHS shouldn't be z-zw. It should be w-wz.

darkphoenix · Dec 2, 2011

sorry that was a typo

SpiralFlex · Dec 2, 2011

$w= \frac{z-2i}{1-z}$

$z=\frac{w+2i}{w+1}$

$2=\frac{|w+2i|}{|w+1|}$

Straight away you will know that this belongs to a family of the circles of Apollonius.

$2|w+1|=|w+2i|$

Let $w=x+iy$

Substituting and expanding,

$2|x+1+yi|=|x+i(y+2)|$

$4(x+1)^2+4y^2=x^2+(y+2)^2$

$3x^2+8x+3y^2-4y=0$

$x^2+\frac{8}{3}x+y^2-\frac{4}{3}y=0$

$(x+\frac{4}{3})^2+(y-\frac{2}{3})^2=\frac{20}{9}$

Radius: $\frac{\sqrt{20}}{3}$

Centre: $(\frac{4}{3}, \frac{2}{3})$

Note: It is good to geometrically interpret what you are actually doing. Graphical software is suggested by Spiral for loci.

darkphoenix · Dec 2, 2011

Oh, thanks spi. I just realise my mistake in the distance formula part... sign can't even do 2u now

Complex no help (1 Viewer)

darkphoenix

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RealiseNothing

what is that?It is Cowpea

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SpiralFlex

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