Thread: Trying to sketch this graph on the complex plane

1. Trying to sketch this graph on the complex plane

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

$\displaystyle |x + iy +5| = 4$

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

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

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

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

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

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

My notes have steps in polar form:

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

...

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

I'm not sure how these were derived.

2. Re: Trying to sketch this graph on the complex plane

It is a circle with radius 4 and centre at $\displaystyle (x,y) = (-5,0)$.

3. Re: Trying to sketch this graph on the complex plane

Originally Posted by sander
Btw. I think your third, fourth, fifith and six line should include $\displaystyle -y^2$ instead of $\displaystyle y^2$.
Why? I didn't square $\displaystyle i$.

4. 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

5. 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 $\displaystyle (x+ 5)^2+ y^2= 16$ 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 $\displaystyle (x- a)^2+ (y- b)^2= r^2$, the equation of a circle with center (a, b) and radius r, with a= -5, b= 0, and r= 4.

6. Re: Trying to sketch this graph on the complex plane

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 $\displaystyle (x+ 5)^2+ y^2= 16$ 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 $\displaystyle (x- a)^2+ (y- b)^2= r^2$, 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