Trying to sketch this graph on the complex plane

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September 10th 2011, 06:17 AM
terrorsquid

Trying to sketch this graph on the complex plane

Sketch the graph of $|z+5| = 4$ ; that is, find all $z \in \mathbb{C}$ which satisfy this equation. What geometric shape is it?

$|x + iy +5| = 4$

$|(x+5) +iy|=4$

$\sqrt{(x+5)^2+y^2} = 4$

$(x+5)^2+y^2=16$

$x^2+10x+25 +y^2 = 16$

$x^2 + y^2 = -10x - 9$

I don't know what to do here. Is it a circle?

My notes have steps in polar form:

$(\sqrt3)^4(cos\left(\frac{8\pi}{3}\right)+isin \left( \frac{8\pi}{3}\right) )$

...

$= 9(-\frac{1}{2}+i\frac{\sqrt3}{2})$

I'm not sure how these were derived.
September 10th 2011, 06:24 AM
sander

Re: Trying to sketch this graph on the complex plane

It is a circle with radius 4 and centre at $(x,y) = (-5,0)$ .
September 10th 2011, 06:29 AM
terrorsquid

Re: Trying to sketch this graph on the complex plane

Quote:

Originally Posted by sander

Btw. I think your third, fourth, fifith and six line should include $-y^2$ instead of $y^2$ .

Why? I didn't square $i$ .
September 11th 2011, 01:35 AM
andrew2322

Re: Trying to sketch this graph on the complex plane

because y is the imaginary part, not iy, therefore you dont square i along with y.

it is a circle with radius 4 and center at (-5,0) as Sander pointed out
September 11th 2011, 04:34 AM
HallsofIvy

Re: Trying to sketch this graph on the complex plane

Sander and andrew2322 are using the fact that |z- a| is the distance from the point z to the point a in the complex plane. |z+ 5|= |z- (-5)| so |z+ 5|= 4 is satisfied for all points whose distanced from -5 is equal to 4- in other words, the circle with center -5 and radius 4.

Also, you arrived at $(x+ 5)^2+ y^2= 16$ and then proceeded to multiply out the square, etc. You should not have done that. You should have recognized immediately that this is of the form $(x- a)^2+ (y- b)^2= r^2$ , the equation of a circle with center (a, b) and radius r, with a= -5, b= 0, and r= 4.
September 11th 2011, 08:33 AM
terrorsquid

Re: Trying to sketch this graph on the complex plane

Quote:

Originally Posted by HallsofIvy

Sander and andrew2322 are using the fact that |z- a| is the distance from the point z to the point a in the complex plane. |z+ 5|= |z- (-5)| so |z+ 5|= 4 is satisfied for all points whose distanced from -5 is equal to 4- in other words, the circle with center -5 and radius 4.

Also, you arrived at $(x+ 5)^2+ y^2= 16$ and then proceeded to multiply out the square, etc. You should not have done that. You should have recognized immediately that this is of the form $(x- a)^2+ (y- b)^2= r^2$ , the equation of a circle with center (a, b) and radius r, with a= -5, b= 0, and r= 4.

That made a lot of sense, thanks! Yah, I didn't realise it was a circle. I only knew it in the form x^2 + y^2 = r^2