How do i find $z$ with $|z+9|=3|z+1|$?

Question

When z is a complex number, how do I find $z$ with $|z+9|=3|z+1|$? I tried replacing $z$ with $x + yi$ but I'm stuck with all of them under the square root.

Answer 1

Let $z=x+yi,$ where $\{x,y\}\subset\mathbb R$.

Thus, we have $$\sqrt{(x+9)^2+y^2}=3\sqrt{(x+1)^2+y^2}$$ or $$(x+9)^2+y^2=9((x+1)^2+y^2),$$ which gives an equation of the circle $$x^2+y^2=9.$$

Answer 2

Hint:  the locus of points in the plane with constant ratio $\,\left|\dfrac{z+9}{z+1}\right| = 3\,$ between the distances to the two fixed points $\,-9, -1\,$ is a circle of Apollonius having as diameter the points that divide the segment between $\,-9\,$ and $\,-1\,$ in ratio $\,3:1\,$ internally and respectively externally. See this answer for the complete derivation.

Answer 3

If you want, you can do this entirely without splitting $z$ into $x+iy$. We have $$\eqalign{|z+9|=3|z+1|\quad &\Leftrightarrow\quad |z+9|^2=9|z+1|^2\cr &\Leftrightarrow\quad (z+9)(\overline z+9)=9(z+1)(\overline z+1)\cr &\Leftrightarrow\quad \cdots\quad \hbox{(a bit of easy algebra, leave it up to you)}\cr &\Leftrightarrow\quad z\overline z=9\cr &\Leftrightarrow\quad |z|=3\cr}$$ which is a circle with centre $0$, radius $3$.