complex numbers (1 Viewer)

maths lover · Nov 27, 2011

i am having troubles sketching the locus of this inequality to find the range of modulus. the question is modulus of(z-sqrt(3)-i)=2. i might be brain dead or something but why does this graph cut at the origin, thanks.

SpiralFlex · Nov 27, 2011

Let us interpret this result geometrically then algebraically,

$|z-(\sqrt{3}+i)|=2$

So geometrically, we know this is a circle with a radius of 2. What this means is how far is the locus of z from the point $(\sqrt 3, i)$ (Centre)

So let us draw our pretty diagram,

And there you have it.

Algebraically we let,

$z=x+iy$

$|x+iy-(\sqrt{3}+i)|=2$

$|x-\sqrt{3}+i(y-1)|=2$

$\sqrt{(x-\sqrt{3})^2+(y-1)^2}=2$

$(x-\sqrt{3})^2+(y-1)^2=4$

maths lover · Nov 27, 2011

yeah i did all that i just dont get why it crosses the origin. sorry.
EDIT: i understand that you can use Pythagoras to see that it cuts the origin, but how at first glance did you know it cut the origin.

SpiralFlex · Nov 27, 2011

For the locus of z to cross the origin.

$(x-\sqrt{3})^2+(y-1)^2=4$

Has to be true for point $(0,0)$

$(0-\sqrt{3})^2+(0-1)^2=4$

$3+1=4$

Ah, it's true. Therefore it crosses the origin.

If you have a curve that is close to the origin and you do not know if it crosses it, there is no harm in testing. Only takes 1 micro-second.

maths lover · Nov 27, 2011

oh ok i was brain dead, since i didnt find the equation of the circle just geometrically interpreted and ended up drawing it so that it didnt cut the origin. thanks spiral for everything.

complex numbers (1 Viewer)

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