Displaying $ |z+4|=3|z|$ in an Argand diagram

Question

Identify in an Argand diagram the points corresponding to: $$|z+4|=3|z|$$ I am unsure how to approach this question and would appreciate any help anyone can provide with the answer and the intuition behind the method. Thanks.

Answer 1

Hint: By substituting $z=x+iy$ we get \begin{align} |x+4+yi|=3|x+yi| & \implies \sqrt{(x+4)^2+y^2}=3\sqrt{x^2+y^2} \\ & \implies (x + 4)^2 + y^2 = 9(x^2 + y^2) \\ & \implies (x + 4)^2 - 9x^2 = 8y^2 \\ & \implies y = \pm \sqrt{\frac{(x + 4)^2 - 9x^2}{8}} =\pm \frac{\sqrt{-8x^2 + 16x + 16}}{2 \sqrt{2}} \end{align}

Answer 2

Have look at this question:

Intuitively, why should I expect a circle in the complex plane from the equation $\left|\frac{z-1}{z+1}\right| = c$?

Rewrite your equation $\dfrac{|z+4|}{|z-0|}=3$

The solution to be expected is a circle cutting the line $(AO)$ where $\begin{cases}z_A=-4 &:& A(-4,0)\\z_0=0 &:& O(0,0)\end{cases}$.

The diameter $[IJ]$ is given by the centroids $\begin{cases}\vec{IA}-3\vec{IO}=\vec 0\\\vec{JA}+3\vec{JO}=\vec 0\end{cases}\iff\begin{cases}I=-\frac A2=(2,0)\\J=\frac A4=(-1,0)\end{cases}$

Answer 3

If $z=x+yi$, with real $x,y$ then: $$(x + 4)^2 + y^2 = 9(x^2 + y^2)$$

so we have $$x^2-x-2 +y^2=0\;\;\;/+{1\over 4}$$

$$(x-{1\over 2})^2+y^2 = {9\over 4}$$

so it is a circle with center at $({1\over 2},0)$ and $r= {3\over 2}$

Answer 4

If $|z+4|=3|z|$, and we let $z=x+\mathrm iy$, we get $$|x+\mathrm i y + 4| = 3|x+\mathrm i y|$$ $$|(x+4)+\mathrm iy|=3|x+\mathrm i y|$$ $$\sqrt{(x+4)^2+y^2}=3\sqrt{x^2+y^2}$$ $$(x+4)^2+y^2 = 9(x^2+y^2)$$ $$x^2+8x+16+y^2=9x^2+9y^2$$ $$0=8x^2-8x+8y^2-16$$ $$0=x^2-x+y^2-2$$ $$0=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+y^2-2$$ $$\frac{9}{4} = \left(x-\frac{1}{2}\right)^2+y^2$$

This is a circle, centre $(1/2,0)$ with radius $3/2$, i.e.

$$\left|z-\frac{1}{2}\right|=\frac{3}{2}$$